Permutations & Combinations, Class 11 Mathematics R.D Sharma Question Answer

Page. 16.4 Ex 16.1

Q1.

Answer :
(i) 30!28! = 30×29×28!28! ∵ n!= n(n-1)! =30×29 = 870

(ii) 11!-10!9!=11×10×9! – 10×9!9! ∵ n! =n(n-1)! =9!(110-10)9! =100

(iii) LCM of (6!,7! and 8!):
n! = n(n-1)!
Therefore, (6!,7! and 8!) can be rewritten as:
8! = 8×7×6!
7! = 7×6!
6! = 6!
∴ LCM of (6!,7! and 8!) = LCM [8×7×6!, 7×6! , 6!] = 8×7×6! = 8!

Q2.

Answer :

LHS=19!+110!+111! =19!+110×9!+111×10×9! =110+11+111×10×9! =12211! = RHS Hence, proved.

Q3.

Answer :

(i) 14!+15! = x6!⇒14!+15(4!)= x6!⇒5+15(4!) = x6!⇒ 65!= x6!⇒ 65!= x6×5!⇒x = 36

(ii) x10! =18!+19!⇒x10! =18!+19(8!)⇒x10! =9+19(8!) ⇒x10! = 109!⇒x10×9! = 109!⇒x=100

(iii) 16!+17!=x8!⇒16!+17(6!)=x8!⇒7+17(6!)=x8!⇒ 87!=x8!⇒ 87!=x8×7!⇒x = 64

Q4.

Answer :

i 5×6×7×8×9×10=1×2×3×4×5×6×7×8×9×101×2×3×4 =10!4!

ii 3×6×9×12×15×18=3×1×3×2×3×3×3×4×3×5×3×6 =361×2×3×4×5×6 =366!

iii n+1n+2n+3…2n=123…nn+1n+2n+3…2n123…n =2nn!

iv 135………2n-1 =135………2n-1246………2n246………2n =12345………2n-12n2n123………n =2n!2nn!

Q5.

Answer :

(i) LHS = (2 +3)!
= 5!
= 120
RHS = 2! + 3!
= 2 + 6
= 8
Since LHS ≠ RHS,
Thus, (i) is false.

(ii) LHS = (2 × 3)!
= 6!
= 720
RHS = 2! × 3!
= 2 × 6
= 12
LHS ≠ RHS
Thus, (ii) is false.

Q6.

Answer :

RHS = n! + (n + 1)!
= n! + (n + 1)(n!)
= n! ( 1+ n + 1)
= n! (n+2) = LHS
Hence, proved.

Q7.

Answer :

(n + 2)! = 60 [(n − 1)!]
⇒(n + 2)×(n + 1)×(n)×(n − 1)! = 60 [(n − 1)!]
⇒(n + 2)×(n + 1)×(n) = 60
⇒(n + 2)×(n + 1)×(n) = 5×4×3
∴ n = 3

Q8.

Answer :

(n + 1)! = 90 [(n − 1)!]
⇒(n + 1)×(n)×(n−1)! = 90[(n − 1)!]
⇒(n + 1)×(n) = 90
⇒(n + 1)×(n) = 10 × 9
On comparing, we get:
n = 9

Q9.

Answer :

(n + 3)! = 56 [(n + 1)!]
⇒(n + 3)×(n + 2)×(n + 1)! = 56 [(n + 1)!]
⇒(n + 3)×(n + 2) = 56
⇒(n + 3)×(n + 2) = 8×7
⇒n + 3 = 8
∴ n = 5

Q10.

Answer :

(2n)!3!(2n-3)!:n!2!(n-2)!=44:3 ⇒ (2n)!3!(2n-3)!×2!(n-2)!n!=443⇒(2n)(2n-1)(2n-2) [(2n-3)!]3(2!)(2n-3)!×2!(n-2)!n(n-1) [(n-2)!]=443⇒(2n)(2n-1)(2n-2)3×1n(n-1) = 443⇒(2n)(2n-1)(2)(n-1)3×1n(n-1) = 443⇒4(2n-1)n(n-1)3×1n(n-1) = 443⇒4(2n-1)3= 443⇒(2n-1) = 11⇒2n = 12⇒n = 6

Q11.

Answer :

(i) LHS= n!(n-r)! =nn-1n-2n-3n-4 … n-r+1n-r!(n-r)! =nn-1n-2n-3n-4 … n-r+1 =nn-1n-2n-3n-4 … n-r-1 = RHS

(ii) LHS = n!n-r!r!+n!n-r+1! =n!n-r!r!+ n!(n-r+1) [(n-r)!] =n!n-r+1+ n!r!r!n-r+1 [(n-r)!] =n!n+1-n!r! + n!r!r!n-r+1n-r! =n!(n+1)r!n-r+1n-r! =n+1!r!n-r+1! = RHS Hence proved.

Q12.

Answer :

LHS = 2n +1!n! =2n+12n2n-1….4321n! =135………2n-12n+1246………2nn! =2n135………2n-12n+1123………nn! =2n135………2n-12n+1n!n! =2n135………2n-12n+1 = RHS Hence, proved.

Page. 16.14 Ex 16.2

Q1.

Answer :

No. of boys in the class = 27
No. of girls in the class = 14
Ways to select a boy = 27
Similarly, ways to select a girl = 14
∴ Number of ways to select 1 boy and 1 girl = 27× 14 = 378

Q2.

Answer :

Number of fountain pen varieties = 10
Number of ball pen varieties = 12
Number of pencil varieties = 5
Ways to select a fountain pen = 10
Ways to select a ball pen = 12
Ways to select a pencil = 5

Ways to select a fountain pen, a ball pen and a pencil = 10 ×12 × 5 = 600 (Using the fundamental principle of multiplication)

Q3.

Answer :

Number of routes from Goa to Bombay = 2
Number of routes from Bombay to Delhi = 3
Using fundamental principle of multiplication:
Number of routes from Goa to Delhi via Bombay = 2 × 3 = 6

Q4.

Answer :

The first day of the year can be any one of the days of the week, i.e the first day can be selected in 7 ways.
But, the year could also be a leap year.
So, the mint should prepare 7 calendars for the non-leap year and 7 calendars for the leap year.
So, total number of calendars that should be made = 7 + 7 = 14

Q5.

Answer :

Number of ways of sending 1 parcel via registered post = 5
Number of ways of sending 4 parcels via registered post through 5 post offices = 5×5×5×5 = 625

Q6.

Answer :

Number of outcomes when the coin is tossed for the first time = 2
Number of outcomes when the coin is tossed for the second time = 2
Thus, there would be 2 outcomes, each time the coin is tossed.
Total number of possible outcomes on tossing the coin five times = 2×2×2×2×2 = 32

Q7.

Answer :

Number of ways of answering the first question = 2 (either true or false)
Similarly, each question can be answered in 2 ways.
∴ Total number of ways of answering all the 10 questions = 2×2×2×2×2×2×2×2×2×2 = 210 = 1024

Q8.

Answer :

Number of ways of marking each of the ring = 10 different letters
∴ Total number of ways of marking any letter on these three rings = 10×10×10 = 1000
Out of these 1000 combinations of the lock, 1 combination will be successful.
∴ Total number of unsuccessful attempts = 1000 -1 = 999

Q9.

Answer :

Number of ways of answering the first three questions = 4 each
Number of ways of answering the remaining three questions = 2 each
∴ Total number of ways of answering all the questions = 4×4×4×2×2×2 = 512

Page. 16.15 Ex 16.2

Q10.

Answer :

Number of books on mathematics = 5
Number of books on physics = 6
Number of ways of buying a mathematics book = 5
Similarly, number of ways of buying a physics book = 6
(i) By using fundamental principle of multiplication:
Number of ways of buying a mathematics and a physics book = 6×5 = 30
(ii) By using the fundamental principle of addition:
Number of ways of buying either a mathematics or a physics book = 6 + 5 = 11

Q11.

Answer :

Number of flags = 7
∴ Number of ways of selecting one flag = 7
Number of ways of selecting the other flag = 6 (as only 6 colours are available for use)
A signal requires use of two flags
∴ Total number of signal that can be generated = 7×6 = 42

Q12.

Answer :

A boy can be selected from the first team in 6 ways and from the second team in 5 ways.
∴ Number of ways of arranging a match between the boys of the two teams = 6×5 = 30
Similarly, A girl can be selected from the first team in 4 ways and from the second team in 3 ways.
∴ Number of ways of arranging a match between the girls of the two teams = 4×3= 12
∴ Total number of matches = 30 + 12 = 42

Q13.

Answer :

Number of competitors in the race = 12
Number of competitors who can come first in the race = 12
Number of competitors who can come second in the race = 11 (as one competitor has already come first in the race)
Number of competitors who can come third in the race = 10
∴ Total number of ways of awarding the first three prizes = 12×11×10 = 1320

Q14.

Answer :

Number of ways of selecting the first term from the set {1, 2, 3} = 3
Corresponding to each of the selected first terms, the number of ways of selecting the common difference from the set {1, 2, 3, 4, 5} = 5
∴ Total number of AP’s that can be formed = 3×5 = 15

Q15.

Answer :

Total number of teachers in the college = 36
Number of ways of selecting a principal = 36
Number of ways of selecting a vice-principal = 35 (as one of the teacher is already being selected for the post of principal)
Similarly, number of ways of selecting the teacher-incharge = 34
∴ Total number of ways of selecting all the three = 36×35×34 = 42840

Q16.

Answer :

The thousand’s place cannot be zero.
∴ Number of ways of selecting the thousand’s digit = 9
Number of ways of selecting the ten’s digit = 9 ( as repetition of digits is not allowed and one digit has already been used in the thousand’s place)
Similarly, number of ways of selecting the unit’s digit = 8 (as two digits have been used for the thousand’s and the ten’s places)
∴ Total three digit number that can be formed = 9×9×8 = 648

Q17.

Answer :

Available digits for filling any place = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}
Since the thousand’s place cannot be zero, available digits to fill the thousand’s place = 9
Number of ways of filling the ten’s digit = 10
Similarly, number of ways of filling the unit’s digit = 10
∴ Total number of three digit numbers = 9×10×10 = 900

Q18.

Answer :

Available digits for filling any place = {1, 2, 3, 4, 5, 6, 7, 8, 9, 0}
Since the thousand’s place cannot be zero, the number of ways of filling the thousand’s place is 9.
Number of ways of filling the ten’s place = 10
Number of ways of filling the unit’s place = 5 {1, 3, 5, 7, 9}
Total 3-digit odd numbers = 9×10×5 = 450

Q19.

Answer :

(i) Since the first digit cannot be zero, the number of ways of filling the first digit = 9
Number of ways of filling the second digit = 9 (Since repetition is not allowed)
Number of ways of filling the third digit = 8
Number of ways of filling the fourth digit = 7
Number of ways of filling the fifth digit = 6
Total number of licence plates that can be made = 9×9×8×7×6 = 27216

(ii) Since the first digit cannot be zero, the number of ways of filling the first digit = 9
Number of ways of filling the second digit = 10 (Since repetition is allowed)
Number of ways of filling the third digit = 10
Number of ways of filling the fourth digit = 10
Number of ways of filling the fifth digit = 10
Total number of licence plates that can be made = 9×10×10×10×10 = 90000

Q20.

Answer :

Since the number has to be greater than 7000, the thousand’s place can only be filled by three digits, i.e. 7, 8 and 9.
Now, the hundred’s place can be filled with the remaining 4 digits as the repetition of the digits is not allowed.
The ten’s place can be filled with the remaining 3 digits.
The unit’s place can be filled with the remaining 2 digits.
Total numbers that can be formed = 3×4×3×2 = 72

Q21.

Answer :

Since the number has to be greater than 8000, the thousand’s place can be filled by only two digits, i.e. 8 and 9.
Now, the hundred’s place can be filled with the remaining 4 digits as the repetition of the digits is not allowed.
The ten’s place can be filled with the remaining 3 digits.
The unit’s place can be filled with the remaining 2 digits.
Total numbers that can be formed = 2×4×3×2 = 48

Q22.

Answer :

Number of seats available to the first person = 6
Number of seats available to the second person = 5
Number of seats available to the third person = 4
Number of seats available to the fourth person = 3
Number of seats available to the fifth person = 2
Number of seats available to the sixth person = 1
Total number of ways of making the seating arrangement = 6×5×4×3×2×1 = 720

Q23.

Answer :

Since the first digit cannot be zero, number of ways of filling the first digit = 9
Number of ways of filling the second digit = 9 (as repetition is not allowed or the digits are distinct)
Number of ways of filling the third digit = 8
Number of ways of filling the fourth digit = 7
Number of ways of filling the fifth digit = 6
Number of ways of filling the sixth digit = 5
Number of ways of filling the seventh digit = 4
Number of ways of filling the eighth digit = 3
Number of ways of filling the ninth digit = 2
Total such 9-digit numbers = 9×9×8×7×6×5×4×3×2 = 9 (9!)

Q24.

Answer :

Since the number is less than 1000, it could be a three-digit, two-digit or single-digit number.
Case I: Three-digit number:
Now, the hundred’s place cannot be zero. Thus, it can be filled with three digits, i.e. 3, 5 and 7.
Also, the unit’s place cannot be zero. This is because it is an odd number and one digit has already been used to fill the hundred’s place.
Thus, the unit’s place can be filled by only 2 digits.
Number of ways of filling the ten’s digit = 2 (as repetition is not allowed)
Total three-digit numbers that can be formed = 3×2×2 = 12

Case II: Two-digit number:
Now, the ten’s place cannot be zero. Thus, it can be filled with three digits, i.e. 3, 5 and 7.
Also, the unit’s place cannot be zero. This is because it is an odd number and one digit has already been used to fill the ten’s place,
Thus, the unit’s place can be filled by only 2 digits.
Total two-digit numbers that can be formed = 3×2 = 6

Case III: Single-digit number:
It could be 3, 5 and 7.
Total single-digit numbers that can be formed = 3
Hence, required number = 12 + 6 + 3 = 21

Q25.

Answer :

The hundred’s place can be filled by {1, 3, 5, 7, 9), i.e. 5 digits.
The ten’s place can now be filled by 4 digits (as one digit is already used in the hundred’s place and repetition is not allowed )
Similarly, the unit’s place can be filled by 3 digits.
Total number of 3-digit numbers = 5×4×3 = 60

Q26.

Answer :

Number of ways of filling the first digit = 6
Number of ways of filling the second digit = 5 (as repetition is not allowed)
Number of ways of filling the third digit = 4
Number of ways of filling the fourth digit =3
Number of ways of filling the fifth digit = 2
Number of ways of filling the sixth digit = 1
Total numbers = 6×5×4×3×2×1 = 720

Q27.

Answer :

The first digit cannot be zero. Thus, the first digit can be filled in 5 ways.
Number of ways for filling the second digit = 5 (as repetition of digits is not allowed)
Number of ways for filling the third digit = 4
Number of ways for filling the fourth digit = 3
Number of ways for filling the fifth digit = 2
Number of ways for filling the sixth digit = 1
Total numbers = 5×5×4×3×2×1 = 600

Q28.

Answer :

As the number has to be greater than 5000, the first digit can either be 5 or 9.
Hence, it can be filled only in two ways.
Number of ways for filling the second digit = 4
Number of ways for filling the third digit = 3 (as repetition is not allowed)
Number of ways for filling the fourth digit = 2
Total numbers = 2×4×3×2 = 48

Q29.

Answer :

Number of ways of selecting the first letter = 6
Number of ways of selecting the second letter = 5 (as repetition of letters is not allowed)
Number of ways of selecting the digit in the third place = 10
Number of ways of selecting the digit in the fourth place = 9 (as repetition of digits is not allowed)
Number of ways of selecting the digit in the fifth place = 8
Number of ways of selecting the digit in the sixth place = 7
Possible serial numbers = 6×5×10×9×8×7 = 151200

Q30.

Answer :

The digits in the sequence do not repeat.
Number of ways of selecting the first digit = 10
Number of ways of selecting the second digit = 9
Number of ways of selecting the third digit = 8
Total number of possible sequences = 10×9×8 = 720
Of all the possible sequences, only one sequence is successful.
∴ Number of unsuccessful sequences = 720 – 1 = 719

Q31.

Answer :

Assuming that the code of an ATM has all distinct digits.
Number of ways for selecting the first digit = 4
Number of ways for selecting the second digit = 3
Number of ways for selecting the third digit = 2
Number of ways for selecting the fourth digit = 1
Total number of possible codes for the ATM = 4×3×2×1 = 24

Page. 16.16 Ex 16.2

Q32.

Answer :

Number of ways of assigning a job to person A = 3
Number of ways of assigning the remaining jobs to person B = 2 (since one job has already been assigned to person A)

The number of ways of assigning the remaining job to person C = 1
Total number of ways of job assignment = 3×2×1 = 6

Q33.

Answer :

Case I: Four-digit number
Total number of ways in which the 4 digit number can be formed = 4×4×4×4 = 256

Now, the number of ways in which the 4-digit numbers greater than 4321 can be formed is as follows:
Suppose, the thousand’s digit is 4 and hundred’s digit is either 3 or 4.
∴ Number of ways = 2×4×4 = 32
But 4311, 4312, 4313, 4314, 4321 (i.e. 5 numbers) are less than or equal to 4321.
∴ Remaining number of ways = 256-32-5=229

Case II: Three-digit number
The hundred’s digit can be filled in 4 ways.
Similarly, the ten’s digit and the unit’s digit can also be filled in 4 ways each. This is because the repetition of digits is allowed.
∴ Total number of three-digit number = 4×4×4 = 64

Case III: Two-digit number
The ten’s digit and the unit’s digit can be filled in 4 ways each. This is because the repetition of digits is allowed.
∴ Total number of two digit numbers = 4×4 = 16

Case IV: One-digit number
Single digit number can only be four.
∴ Required numbers = 229 + 64 + 16 +4 = 313

Q34.

Answer :

The first digit of the number cannot be zero. Thus, it can be filled in 5 ways.
The number of ways of filling the second digit = 5 (as the repetition of digits is not allowed)
The number of ways of filling the third digit = 4
The number of ways of filling the fourth digit = 3
The number of ways of filling the fifth digit = 2
The number of ways of filling the sixth digit = 1
∴ Required numbers = 5×5×4×3×2×1 = 600

For the number to be divisible by 10, the sixth digit has to be zero.
Now, the first digit can be filled in 5 ways.
Number of ways of filling the second digit = 4
Number of ways of filling the third digit = 3
Number of ways of filling the fourth digit = 2
Number of ways of filling the fifth digit = 1
Number of ways of filling the sixth digit = 1
Total numbers divisible by 10 = 5×4×3×2×1×1 = 120

Q35.

Answer :

Number of possible outcomes on one dice = 6 {1,2,3,4,5,6}
Number of possible outcomes on both the other two dice = 6
∴ Total number of outcomes when three dice are thrown = 6×6×6 = 216

Q36.

Answer :

Total number of outcomes when a coin is tossed once = 2 (Heads, Tails)
Number of outcomes when the coin is tossed for the second time = 2
∴ Number of outcomes when the coin is tossed thrice = 2×2×2 = 8
Similarly, the number of outcomes when the coin is tossed four times = 2×2×2×2 = 16
Similarly, the number of outcomes when the coin is tossed five times = 2×2×2×2×2 = 32
Similarly, the number of outcomes when the coin is tossed ‘n’ times = 2×2×…..n times = 2n

Q37.

Answer :

Each of the toy can be distributed in 5 ways.
∴ Total number of ways of distributing the toys = 5×5×5×5×5×5×5×5 = 58

Q38.

Answer :

Each of the 5 letters can be posted in any one of the 7 letter boxes.
∴ Required number of ways of posting the letters = 7×7×7×7×7 = 75

Q39.

Answer :

Required number of possible outcomes = (Total number of outcomes – Number of possible outcomes in which 5 does not appear on any of the dice.)
Total number of outcomes when a single dice is rolled = 6
∴ Total number of outcomes when two dice are rolled = 6×6
Similarly, total number of outcomes when three dice are rolled = 6×6×6 = 216
Number of possible outcomes in which 5 does not appear on any dice = 5×5×5 = 125
∴ Required number of possible outcomes = 216 – 125 = 91

Q40.

Answer :

Any one of the twenty balls can be put in the first box. Thus, there are twenty different ways for this.
Now, remaining 19 balls are to be put into the remaining 4 boxes. This can be done in 419 ways because there are four choices for each ball.
∴ Required number of ways = 20×419

Q41.

Answer :

Each ball can be distributed in 3 ways.
∴ Required number of ways in order to distribute 5 balls = 3×3×3×3×3 = 243

Q42.

Answer :

Each letter can be posted in any one of the 4 letter boxes.
Number of ways of posting one letter = 4
∴ Required number of ways of posting the 7 letters = 4×4×4×4×4×4×4 = 47

Q43.

Answer :

(i) Since no student gets more than one prize; the first prize can be given to any one of the five students.
The second prize can be given to anyone of the remaining 4 students.
Similarly, the third prize can be given to any one of the remaining 3 students.
The last prize can be given to any one of the remaining 2 students.
∴ Required number of ways = 5×4×3×2 = 5!

(ii) Since a student may get any number of prizes, the first prize can be given to any of the five students. Similarly, the rest of the three prizes can be given to the each of the remaining 4 students.
∴ Required number of ways = 5×5×5×5 = 625

(iii) None of the students gets all the prizes.
∴ Required number of ways = {Total ways of distributing the prizes in a condition wherein a student may get any number of prizes – Total ways in a condition in which a student receives all the prizes} =
625 – 5 = 620

Q44.

Answer :

The thousand’s place can be filled by any of the 5 digits.
∴ Number of ways of filling the thousand’s place = 5
Since the digits can repeat in the number, the hundred’s place, the ten’s place and the unit’s place can each be filled in 5 ways.
∴ Total numbers = 5×5×5×5 = 625

Q45.

Answer :

Since the hundred’s place cannot be zero, it can be filled by any of the 4 digits (1, 3, 5 and 7).
∴ Number of ways of filling the hundred’s place = 4
Since the digits can be repeated in the number, the ten’s place and the unit’s place can each be filled in 5 ways.
∴ Total numbers = 4×5×5 = 100

Q46.

Answer :

Since the number is less than 1000, it means that it is a three-digit number, a two-digit number or a single-digit number.
Three-digit numbers:
The hundred’s place can be filled by 5 digits neglecting zero as it can’t be zero.
The ten’s place and the unit’s place can be filled by 6 digits.
So, total number of three digit numbers = 5×6×6 = 180

Two-digit numbers:
The ten’s place can be filled by 5 digits, except zero.

The unit’s digit can be filled by 6 digits.
Total two digit numbers = 5×6 = 30

Single digit numbers are 1, 2, 3, 4, 5 as 0 is not a natural number. Thus, on neglecting it, we get 5 numbers.
Total required numbers = 180 + 30 + 5 = 215

Q47.

Answer :

Total outcomes when a coin is tossed = 2 (head or tail)
When a coin is tossed three times:
Total number of possible outcomes = 2×2×2 = 8

Q48.

Answer :

The first two digits are fixed as 67.
As repetition of digits in not allowed, the number of available digits to fill the remaining places is 8.
The third place can be filled in 8 ways.
The fourth place can be filled in 7 ways.
The fifth place can be filled in 6 ways.
Total number of such telephone numbers = 8×7×6 = 336

Page. 16.28 Ex 16.3

Q1.

Answer :

(i) ⁸P₃
nPr = n!(n-r)!

∴ ⁸P₃ =8!(8-3)!

= 8!5!

=8(7)(6)(5!)5!=8×7×6 = 336

ii 10P4=10!(10-4)! = 10!6! =10(9)(8)(7)(6!)6! =10×9×8×7 = 5040

iii 6P6=6!(6-6)! = 6!0! = 6!1 (Since , 0! = 1) = 720

(iv) P(6,4)
It can also be written as ⁶P₄ .
6P4 = 6!2! = 6(5)(4)(3)(2!)2! =6×5×4×3 = 360

Q2.

Answer :

P (5, r) = P (6, r − 1)
or ⁵P_r = ⁶P_r-1
5!5-r!=6!6-r+1!⇒6-r+1!5-r!=6!5!⇒(7-r)!5-r!=65!5!⇒7-r6-r5-r!5-r!=6⇒7-r6-r= 6⇒7-r6-r =3×2On comparing the above two equations, we get:7-r=3⇒ r= 4

Q3.

Answer :

5 P(4, n) = 6. P (5, n − 1)
5 ⁴P_n = 6⁵P_n-1
⇒5×4!4-n!=6×5!5-n+1!⇒5×6-n!4-n!=6×5!4!⇒5×6-n6-n-16-n-2!4-n=6×5×4!4!⇒5×6-n5-n4-n!4-n=6×5⇒6-n5-n = 6⇒6-n5-n = 3×2On comparing the LHS and the RHS, we get:⇒6-n =3⇒n=3

Page. 16.36 Ex 16.4

Q1.

Answer :

There are 7 letters in the word FAILURE.
We wish to find the total number of arrangements of these 7 letters so that the consonants occupy only odd positions.
There are 3 consonants and 4 odd positions. These 3 consonants can be arranged in the 4 positions in 4! ways.
Now, the remaining 4 vowels can be arranged in the remaining 4 positions in 4! ways.

By fundamental principle of counting:
Total number of arrangements = 4!×4! = 576

Q2.

Answer :

(i) Number of vowels = 2
Number of consonants = 5
Considering the two vowels as a single entity, we are now to arrange 6 entities taken all at a time.
Total number of ways = 6!
Also, the two vowels can be mutually arranged amongst themselves in 2! ways.

By fundamental principle of counting:
Total number of words that can be formed = 6!×2! = 1440

(ii) Total number of words that can be made with the letters of the word STRANGE = 7! = 5040
Number of words in which vowels always come together = 1440
∴ Number of words in which vowels do not come together = 5040 -1440 = 3600

(iii) There are 7 letters in the word STRANGE.
We wish to find the total number of arrangements of these 7 letters so that the vowels occupy only odd positions.
There are 2 vowels and 4 odd positions.
These 2 vowels can be arranged in the 4 positions in 4×3 ways, i.e. 12 ways.
The remaining 5 consonants can be arranged in the remaining 5 positions in 5! ways.
By fundamental principle of counting:
Total number of arrangements = 12×5! = 1440

Q3.

Answer :

Total number of words that can be formed with the letters of the word SUNDAY = 6! = 720
Fixing the first letter as D:
Number of arrangements of the remaining 5 letters, taken 5 at a time = 5! = 120
Number of words with the starting letter D = 120

Q4.

Answer :

There are 8 letters in the word ORIENTAL.
We wish to find the total number of arrangements of these 8 letters so that the vowels occupy only odd positions.
There are 4 vowels and 4 odd positions.
These 4 vowels can be arranged in the 4 positions in 4! ways.
Now, the remaining 4 consonants can be arranged in the remaining 4 positions in 4! ways.
By fundamental principle of counting:
Total number of arrangements = 4!×4! = 576

Q5.

Answer :

Total number of words that can be formed with the letters of the word SUNDAY = 6! = 720
Now, if we fix the first letter as N, the remaining 5 places can be filled with the remaining 5 letters in 5! ways, i.e. 120.
If we fix the first letter as N and the last word as Y:
Remaining 4 places can be filled with 4 letters in 4! ways = 24

Q6.

Answer :

The word GANESHPURI consists of 10 distinct letters.
Number of letters = 10!

(i) If we fix the first letter as G, the remaining 9 letters can be arranged in 9! ways to form the words.
∴ Number of words starting with the letter G = 9!

(ii) If we fix the first letter as P and the last letter as I, the remaining 8 letters can be arranged in 8! ways to form the words.
∴ Number of words that start with P and end with I = 8!

(iii) The word GANESHPURI consists of 4 vowels. If we keep all the vowels together, we have to consider them as a single entity.
So, we are left with the remaining 6 consonants and all the vowels that are taken together as a single entity. This gives us a total of 7 entities that can be arranged in 7! ways.
Also, the 4 vowels can be arranged in 4! ways amongst themselves.

By fundamental principle of counting:
Total number of arrangements = 7!×4! words.

(iv) The word GANESHPURI consists of 4 vowels that have to be arranged in the 5 even places. This can be done in 5! ways.
Now, the remaining 6 consonants can be arranged in the remaining 6 places in 6! ways.
Total number of words in which the vowels occupy even places = 5!×6!

Q7.

Answer :

(i) The word VOWELS consists of 6 distinct letters that can be arranged amongst themselves in 6! ways.
∴ Number of words that can be formed with the letters of the word VOWELS, without any restriction = 6! = 720

(ii) If we fix the first letter as E, the remaining 5 letters can be arranged in 5! ways to form the words.
∴ Number of words starting with the E = 5! = 120

(iii) If we fix the first letter as O and the last letter as L, the remaining 4 letters can be arranged in 4! ways to form the words.
∴ Number of words that start with O and end with L = 4! = 24

(iv) The word VOWELS consists of 2 vowels.
If we keep all the vowels together, we have to consider them as a single entity.
Now, we are left with the 4 consonants and all the vowels that are taken together as a single entity.
This gives us a total of 5 entities that can be arranged in 5! ways.
It is also to be considered that the 2 vowels can be arranged in 2! ways amongst themselves.

By fundamental principle of counting:
∴ Total number of arrangements = 5!×2! = 240

(v) The word VOWELS consists of 4 consonants.
If we keep all the consonants together, we have to consider them as a single entity.
Now, we are left with the 2 vowels and all the consonants that are taken together as a single entity.
This gives us a total of 3 entities that can be arranged in 3! ways.
It is also to be considered that the 4 consonants can be arranged in 4! ways amongst themselves.

By fundamental principle of counting:
∴ Total number of arrangements = 3!×4! = 144

Q8.

Answer :

The word ARTICLE consists of 3 vowels, which have to be arranged in 3 even places. This can be done in 3! ways.
Now, the remaining 4 consonants can be arranged in the remaining 4 places in 4! ways.
∴ Total number of words in which the vowels occupy only even places = 3!×4! = 144

Q9.

Answer :

We arrange any 2 men in ⁷P₂ ways and then the wives of the remaining 5 men can be arranged in ⁵P₂ ways. This is because these two men should not be with their respective wives.
∴ By fundamental principle of counting, the required number of ways = ⁷P₂ × ⁵P₂
=7!5!×5!3! = 7!3! = 840

Q10.

Answer :

‘m’ men can be seated in a row in m! ways.
‘m’ men will generate (m+1) gaps that are to be filled by ‘n’ women = Number of arrangements of (m+1) gaps, taken ‘n’ at a time = ^m=1P_n = m+1!m+1-n!
∴ By fundamental principle of counting, total number of ways in which they can be arranged = m!m+1!m-n+1!

Q11.

Answer :

(i) The word MONDAY consists of 6 letters.
Number of words formed using 4 letters = Number of arrangements of 6 letters, taken 4 at a time = ⁶P₄ = 6!2!= 6×5×4×3 = 360
(ii) Number of words formed using all the letters = Number of arrangements of 6 letters, taken all at a time = ⁶P₆ = 6! = 720
(iii) The word MONDAY consists of 2 vowels and 4 consonants.
The first letter has to be a vowel, which is to be chosen from the two vowels.
This can be done in two ways. The remaining 5 letters can be arranged in 5! ways to form 6 letter words.
⇒ 2 ×5! = 240

Page. 16.37 Ex 16.4

Q12.

Answer :

The word ORIENTAL consists of 8 letters. In order to make three letter words, we need to permute these 8 letters, taken three at a time.
⇒ ⁸P₃ = 8×7×6 = 336

Page. 16.42 Ex 16.5

Q1.

Answer :

(i) This word consists of 12 letters that include three Ns, two Ds and four Es.
The total number of words is the number of arrangements of 12 things, of which 3 are similar to one kind, 2 are similar to the second kind and 4 are similar to the third kind.
⇒ 12!3!2!4! = 1663200

(ii) This word consists of 12 letters that include two Is, two Ts and three Es.
The total number of words is the number of arrangements of 12 things, of which 2 are similar to one kind, 2 are similar to the second kind and 3 are similar to the third kind.
⇒ 12!2!2!3! = 19958400

(iii) This word consists of 7 letters that include two Rs, and two As.
The total number of words is the number of arrangements of 7 things, of which 2 are similar to one kind and 2 are similar to the second kind.
⇒ 7!2!2! = 1260

(iv) This word consists of 5 letters that include two Is.
The total number of words is the number of arrangements of 5 things, of which 2 are similar to one kind.
⇒ 5!2! = 60

(v) This word consists of 8 letters that include two As.
The total number of words is the number of arrangements of 7 things, of which 2 are similar to one kind.
⇒ 8!2! = 20160

(vi) This word consists of 6 letters that include two Ss.
The total number of words is the number of arrangements of 6 things, of which 2 are similar to one kind.
⇒ 6!2! = 360

(vii) This word consists of 6 letters that include two Ss and two Es.
The total number of words is the number of arrangements of 6 things, of which 2 are similar to one kind and 2 are similar to the second kind.
⇒ 6!2!2! = 180

(viii) This word consists of 9 letters that include three Es and two Ss.
The total number of words is the number of arrangements of 9 things, of which 2 are similar to one kind and 2 are similar to the second kind.
⇒ 9!2!3! = 30240

(ix) This word consists of 14 letters that include three Ns, two Os and two Ts.
The total number of words is the number of arrangements of 14 things, of which 3 are similar to one kind, 2 are similar to the second kind and 2 are similar to the third kind.
⇒ 14!3!2!2! =14!24

Q2.

Answer :

The relative positions of all the vowels and consonants is fixed.
The first letter is a vowel. It can be selected out of the 3 three vowels, of which two are same. So, the vowels can be arranged in selecting 3 things, of which two are of the same kind
⇒ 3!2!
The second, third, fifth and sixth letters are consonants that can be filled by the available 4 consonants in 4! ways.
∴ By fundamental principle of counting, the number of words that can be formed = 4!×3!2! = 72

Q3.

Answer :

The word UNIVERSITY consists of 10 letters that include four vowels of which two are same.
Thus, the vowels can be arranged amongst themselves in 4!2!ways.
Keeping the vowels as a single entity, we are left with 7 letters, which can be arranged in 7! ways.
By fundamental principle of counting, we get,
Number of words = 7!× 4!2! = 60480

Q4.

Answer :

When expanded, a³ b² c⁴ would result in total 9 letters.
This is same as permuting 9 things, of which 3 are similar to the first kind, 2 are similar to the second kind and four are similar to the third kind, i.e. three as , two bs and four cs.
Required number of arrangements = 9!3!2!4! = 1260

Q5.

Answer :

The word PARALLEL consists of 8 letters that include two As and three Ls.
Total number of words that can be formed using the letters of the word PARALLEL = 8!2!3! = 3360
Number of words in which all the Ls come together is equal to the condition if all three Ls are considered as a single entity.
So, we are left with total 6 letters that can be arranged in 6!2! ways (divided by 2! since there are two As), which is equal to 360.
Number of words in which all Ls do not come together = Total number of words – Number of words in which all the Ls come together
= 3360 -360
= 3000

Q6.

Answer :

The word MUMBAI consists of 6 letters taht include two Ms.
When we consider both the Ms as a single entity, we are left with 5 entities that can be arranged in 5! ways.
Total number of words that can be formed with all the Ms together = 5! = 120

Page. 16.43 Ex 16.5

Q7.

Answer :

In a dictionary, the words are listed and ranked in alphabetical order. In the given problem, we need to find the rank of the word LATE.
For finding the number of words starting with A, we have to find the number of arrangements of the remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with E, we have to find the number of arrangements of the remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with L, the next alphabetical letter would be A, followed by E and then T, i.e. LAET.
The next alphabetical word would be LATE.
Number of words after which we reach the word LATE = 3!+3!+1+1 = 14

Q8.

Answer :

In a dictionary, the words are listed and ranked in alphabetical order. In the given problem, we need to find the rank of the word MOTHER.
For finding the number of words starting with E, we have to find the number of arrangements of the remaining 5 letters.
Number of such arrangements = 5!
For finding the number of words starting with H, we have to find the number of arrangements of the remaining 4 letters.
Number of such arrangements = 5!

For finding the number of words starting with M, fixing the next letter as E, we have to find the number of arrangements of the remaining 4 letters, which is 4!.
For finding the number of words starting with M, fixing the next letter as H, we have to find the number of arrangements of the remaining 4 letters, which is 4!.

For finding the number of words starting with M, fixing the second letter as O, and the third letter as E, we have to find the number of arrangements of the remaining 3 letters, which is 3!.
For finding the number of words starting with M, fixing the second letter as O, and the third letter as H, we have to find the number of arrangements of the remaining 3 letters, which is 3!.
For finding the number of words starting with M, fixing the second letter as O, and the third letter as R, we have to find the number of arrangements of the remaining 3 letters, which is 3!.
For finding the number of words starting with M, fixing the second letter as O, the third letter as T, and the fourth letter as E, we have to find the number of arrangements of the remaining 2 letters, which is 2!.

Now, the next word formed would be MOTHER.
Number of words after which we reach the word MOTHER = 5!+5!+4!+4!+3!+3!+3!+2!+1 = 309

Q9.

Answer :

In a dictionary, the words are listed and ranked in alphabetical order. In the given problem, we need to find the rank of the word ‘debac’.
For finding the number of words starting with a, we have to find the number of arrangements of the remaining 4 letters.
Number of such arrangements = 4!
For finding the number of words starting with b, we have to find the number of arrangements of the remaining 4 letters.
Number of such arrangements = 4!
For finding the number of words starting with c, we have to find the number of arrangements of the remaining 4 letters.
Number of such arrangements = 4!
For finding the number of words starting with d, fixing the next letter as a, we have to find the number of arrangements of remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with d, fixing the next letter as b, we have to find the number of arrangements of remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with d, fixing the next letter as c, we have to find the number of arrangements of remaining 3 letters.
Number of such arrangements = 3!
For finding the number of words starting with d, fixing the next letter as e:
First word- deabc
Second word- deacb
Third word- debac

Number of words after which we reach the word debac = 4!+4!+4!+3!+3!+3!+1+1+1 = 93

Q10.

Answer :

There are 4 odd digits (1,3,3 and 1) that are to be arranged in 4 odd places in 4!2!2!ways.
The remaining 3 even digits 2, 2 and 4 can be arranged in 3 even places in 3!2!ways.
By fundamental principle of counting:
Required number of arrangements = 4!2!2!×3!2! = 18

Q11.

Answer :

Six ‘+’ signs can be arranged in a row in 6!6! = 1 way
Now, we are left with seven places in which four different things can be arranged in ⁷P₄ ways.
Since all the four ‘-‘ signs are identical, four ‘-‘ signs can be arranged in P474! ways, i.e. 35 ways.
Number of ways = 1× 35 = 35

Q12.

Answer :

We have to arrange 9 flags, out of which 4 are of one kind (red), 2 are of another kind (white) and 3 are of the third kind (green).
∴ Total number of signals that can be generated with these flags = 9!4!2!3! = 1260

Q13.

Answer :

The given digits are 1, 3, 3, 0.
Total numbers that can be formed with these digits = 4!2!
Now, these numbers also include the numbers in which the thousand’s place is 0.
But, to form a four digit number, this is not possible.
∴ Numbers in which the thousand’s place is fixed as zero = Ways of arranging the remaining digits (1,3 and 3) in three places = 3!2!
∴ Four digit numbers = Total numbers – Numbers in which the thousand’s place is 0
= 4!2!- 3!2! = 9

Q14.

Answer :

The word ARRANGE consists of 7 letters including two Rs and two As, which can be arranged in 7!2!2!ways.
∴ Total number of words that can be formed using the letters of the word ARRANGE = 1260
Number of words in which the two Rs are always together = Considering both Rs as a single entity
= Arrangements of 6 things of which two are same (two As)
= 6!2!
= 360
Number of words in which the two Rs are never together = Total number of words – Number of words in which the two Rs are always together
= 1260-360
= 900

Q15.

Answer :

Numbers greater than 50000 can either have 5 or 9 in the first place and will consist of 5 digits.
Number of arrangements having 5 as the first digit = 4!2!

Number of arrangement having 9 as the first digit = 4!2!
∴ Required arrangements = 4!2!+4!2! = 24

Q16.

Answer :

The word SERIES consists of 6 letters including two Ss and two Es.
The first and the last letters are fixed as S.
Now, the remaining four letters can be arranged in 4!2! ways = 12

Q17.

Answer :

Number of words that only end with I = Number of permutations of the remaining 8 letters, taken all at a time = 8!2!

Number of words that start with M and end with I = Permutations of the remaining 7 letters, taken all at a time =7!2!
Number of words that do not begin with M but end with I = Number of words that only end with I – Number of words that start with M and end with I
= 8!2! – 7!2!
= 17640

Q18.

Answer :

One million (1,000,000) consists of 7 digits.
We have digits 2, 3, 0, 3, 4, 2 and 3.
Numbers formed by arranging all these seven digits = 7!2!3!
But, these numbers also include the numbers whose first digit is 0.
This is invalid as in that case the number would be less than a million.
Total numbers in which the first digit is fixed as 0 = Permutations of the remaining 6 digits = 6!2!3!
Numbers that are greater than 1 million = 7!2!3!-6!2!3! = 360

Q19.

Answer :

The word INTERMEDIATE consists of 12 letters that include two Is, two Ts and three Es.
(i) There are 6 vowels (I, I, E, E, E and A) that are to be arranged in six even places = 6!2!3!= 60
The remaining 6 consonants can be arranged amongst themselves in 6!2! ways, which is equal to 360.
By fundamental principle of counting, the number of words that can be formed = 60×360 = 21600

(ii) The relative positions of all the vowels and consonants is fixed.
Arranging the six vowels at their places, without disturbing their respective places, we can arrange the six vowels in 6!2!3! ways.

Similarly, arranging the remaining 6 consonants at their places, without disturbing their respective places, we can arrange the 6 consonants in 6!2! ways.
By fundamental principle of counting, the number of words that can be formed = 6!2!3!×6!2! = 21600

Q20.

Answer :

Total number of books = 12
∴ Required number of arrangements = Arrangements of 12 things of which each of the 4 different books has three copies = 12!3!3!3!3! = 12!(3!)4

Q21.

Answer :

In a dictionary, the words are arranged in alphabetical order. Therefore, in the given problem, we must consider the words beginning with E, H, I, N, T and Z.

∴ Number of words starting with E = 5! = 120
Number of words starting with H = 5! = 120
Number of words starting with I = 5! = 120
Number of words starting with N = 5! = 120
Number of words starting with T = 5! = 120
Now, the word will start with the letter Z.
After Z, alphabetically, the next letter would be E, which is as per the requirement of the word ZENITH.
After ZE, alphabetically, the next letter would be H, i.e. ZEH. The remaining three letters can be arranged in 3! ways.
Now, the next letter would be I, i.e. ZEI. The remaining three letters can be arranged in 3! ways.
Now, the next letter would be N, which as per the requirement of the word ZENITH.

After ZEN, alphabetically, the next letter would be H, i.e. ZENH. The remaining two letters can be arranged in 2! ways.
The next letter would now be I, i.e. ZENI, which is as per the requirement of the word ZENITH.
H will come after ZENI, which would be followed by T.
The word formed is ZENIHT.
The next word would be ZENITH.
Total number of intermediate words = 5×120 + 3! + 3! + 2! + 1 + 1 = 616

Q22.

Answer :

The word MATHEMATICS consists of 11 letters that include two Ms, two As, and two Ts.
Total number of arrangements of the letters of the word MATHEMATICS = 11!2!2!2!
Number of words in which the first word is fixed as C = Number of arrangements of the remaining 10 letters, of which there are two As, two Ms and two Ts = 10!2!2!2!
Number of words in which the first word is fixed as T = Number of arrangements of the remaining 10 letters, of which there are two As and two Ms = 10!2!2!

Q23.

Answer :

In a dictionary, the words are arranged in the alphabetical order. Thus, in the given problem, we must consider the words beginning with I, I, R, S, T and U.
I will occur at the first place as often as the ways of arranging the remaining 5 letters, when taken all at a time.
Thus, I will occur 5! times.
Similarly, R will occur at the first place the same number of times.
∴ Number of words starting with I = 5!
Number of words starting with R = 5!2!
The word will now start with S, which is as per the requirement of the word SURITI.
Alphabetically, the next letter would be I, i.e. SI. The remaining four letters can be arranged in 4! ways.
Alphabetically, the next letter would now be R, i.e. SR. The remaining four letters can be arranged in 4!2! ways.

Alphabetically, the next letter would now be T, i.e. ST. The remaining four letters can be arranged in 4!2! ways.

Alphabetically, the next letter would now be U, i.e. SU, which is as per the requirement of the word SURITI.
After SU, alphabetically, the third letter would be I, i.e. SUI. Thus, the remaining 3 letters can be arranged in 3! ways.
The next third letter that can come is R, i.e. SUR, which is as per the requirement of the word SURITI.
After SUR, the next letter that will come is I, i.e. SURI, which is as per the requirement of the word SURITI.
The next word arranged in the dictionary will be SURIIT.
Then, the next word will be SURITI.
Rank of the word SURITI in the dictionary = 5! + 5!2! + 4! + 4!2! + 4!2! + 3! + 2 = 236
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Questions

Q24.

Answer :

Number of molecules in a chain = 12
Number of molecules with initials A = 3
Number of molecules with initials C = 3
Number of molecules with initials G = 3
Number of molecules with initials T = 3
Thus, total arrangements of all the molecules in the chain = Number of arrangements of 12 things of which 3 are similar to the first kind, 3 are similar to the second kind, 3 are similar to the third kind and 3 are similar to the fourth kind
= 12!3!3!3!3! = 369600

Q25.

Answer :

Number of red discs = 4
Number of yellow discs = 3
Number of green discs = 2
Total number of discs = 9
Total number of arrangements = Number of arrangements of 9 things of which 4 are similar to the first kind, 3 are similar to the second kind and 2 are similar to the third kind = 9!4!3!2! = 1260

Q26.

Answer :

Numbers greater than a million can be formed when the first digit can be any one out of the given digits 1, 2, 0, 2, 4, 2, 4, except 0.
Number of arrangements of the given digits 1, 2, 0, 2, 4, 2, 4 = Arrangements of 7 things of which 3 are similar to the first kind, and 2 are similar to the second kind = 7!2!3!
But, these arrangements also include the numbers in which the first digit is zero. This will make the number less than a million. So, it needs to be subtracted.
Number where the first digit is zero = Number of arrangements of the remaining 6 digits 1, 2, 2, 4, 2, 4
= 6!2!3!
Numbers greater than 1 million = 7!2!3!-6!2!3! = 360

Q27.

Answer :

The word ASSASSINATION consists of 13 letters including three As, four Ss, two Ns and two Is.
Considering all the Ss are together or as a single letter, we are left with 10 letters. Out of these, there are three As, two Ns and two Is.
Number of words in which all the Ss are together = Permutations of 10 letters of which three are similar to the first kind, two are similar to the second kind and two are similar to the third kind = 10!2!2!3! = 151200

Q28.

Answer :

The word ‘INSTITUTE’ consists of 9 letters including two Is and three Ts.
Total number of words that can be formed of the word INSTITUTE = Number of arrangements of 9 things of which 2 are similar to the first kind and 3 are similar to the second kind = 9!2!3!

Page. 16.45 (Very Short Answers)

Q1.

Answer :

Each of the letter can be posted in anyone of the letter boxes.
This means that every letter can be posted in 5 ways.
∴ Total number of ways of posting 4 letters = 5×5×5×5 = 5⁴

Q2.

Answer :

Number of ways in which the first digit can be filled = Number of digits available for filling it = 2 {1,2} (Since the first one cannot be 0)
Number of ways of filling the remaining four palaces = 3 each (as each place can be filled with either 1, 2 or 0)
By fundamental principle of counting, number of five digit numbers that can be formed = 2×3×3×3×3 = 2×3⁴

Q3.

Answer :

Number of ways to draw water from the 1st tap = Number of women available to draw water = 4
Number of ways to draw water from the 2nd tap = Number of women available to draw water = 3
Number of ways to draw water from the 3rd tap = Number of women available to draw water = 2
Number of ways to draw water from the 4th tap = Number of women available to draw water = 1
∴ Total number of ways = 4×3×2×1 = 4! = 24

Q4.

Answer :

Total number of outcomes when 3 dice are thrown = 6×6×6 = 216
Number of outcomes in which there is an odd number on all the three dice = 3×3×3 = 27
∴ Number of outcomes in which there is an even number at least on one dice = {Total possible outcomes} – {Number of outcomes in which there is an odd number on all the three dice } = 216 – 27 = 189

Q5.

Answer :

The word BANANA consists of 6 letters including three As and two Ns.
Considering both Ns together or as a single letter, we are left with 5 letters including three As.
∴ Number of arrangements of 5 things in which 3 are similar to one kind = 5!3! = 20

Q6.

Answer :

The word LATE consists of four letters, which when arranged alphabetically are A, E, L and T.

Initially, the words will start with A.
Number of words starting with A = Number of arrangements of the remaining 3 letters = 3!

Then, the words starting with E will be arranged alphabetically.
Number of words starting with E = Number of arrangements of the remaining 3 letters = 3!

Now, the words will start with L, which is as per the requirement of the word LATE.
Alphabetically, A will follow the letter L, which is also as per the requirement of the word LATE.

Alphabetically, the third letter will be E and then T.
Word formed = LAET
The next word that will be arranged in the dictionary = LATE

Rank of the word LATE = 3! + 3! + 2 = 14

Q7.

Answer :
The word COMMITTEE consists of 9 letters including two Ms, two Ts and two Es.
Number of words that can be formed out of the letters of the word COMMITTEE
= Number of arrangements of 9 things of which 2 are similar to the first kind,
2 are similar to the second kind and 2 are similar to the third kind = 9!2!2!2!=9!2!3

Q8.

Answer :

The word ‘MATHEMATICS’ consists of 11 letters including two Ms, two Ts and two As
Number of words that can be formed out of the letters of the word MATHEMATICS = Number of arrangements of 11 things of which 2 are similar to the first kind, 2 are similar to the second kind and 2 are similar to the third kind = 11!2!2!2!

Q9.

Answer :

Each of the six men can be arranged amongst themselves in 6! ways.
The five women can be arranged amongst themselves in the six places in 5! ways.
∴ By fundamental principle of counting, total number of ways = 6!×5!

Q10.

Answer :

Five boys can be arranged amongst themselves in 5! ways, at the places shown above.
The three girls are now to be arranged in the remaining four places taken three at a time = ⁴P₃ = 4!
By fundamental principle of counting, total number of ways = 5!×4! = 120×24 = 2880

Q11.

Answer :

Every number after 6! (i.e. 7! onwards) till 200! will consist a power of 2 and 7, which will be exactly divisible by 14.

So, we need to divide only the sum till 6!.

1! + 2! + 3! + 4! + 5! + 6! = 1 + 2 + 6 + 24 + 120 + 720 = 873

When 873 is divided, the remainder would be same as when 1! + 2! + 3! + … + 200! is divided by 14.
Remainder obtained when 1! + 2! + 3! + … + 200! is divided by 14 = Remainder obtained when 873 is divided by 14 = 5

Q12.

Answer :

Disclaimer:- (1) Here, we can form 4 digits, 5 digits , 6 digits numbers and so on…. using the given digits. Thus, infinite numbers can be formed.
(2) Taking into account only four digit numbers.

We have to find all the numbers that can be formed by using the digits 1, 2, 3 and 4. This means that repetition of digits is not allowed as all the digits have to be used.
Total numbers that can be formed = Number of arrangements of four digits, taken all at a time = 4! = 24

Q13.

Answer :

Each of the seven men can be arranged amongst themselves in 7! ways.
The women can be arranged amongst themselves in seven places, in 6! ways (i.e. n things can be arranged in (n-1)! ways around a round table).
By fundamental principle of counting, total number of ways = 7!×6!

Page. 16.45 (Multiple Choice Questions)

Q1.

Answer :

(d) r ! ^n-3C_r-3
Here, we have to permute n things of which 3 things are to be included.
So, only the remaining (n-3) things are left for permutation, taking (r-3) things at a time. This is because 3 things have already been included.
But, these r things can be arranged in r! ways.
∴ Total number of permutations = r ! ^n-3C_r-3

Q2.

Answer :

(c) 7200

Total number of vowels = 4
Total number of consonants = 5

Number of selection of 2 vowels taking from 4 vowels
= ⁴C₂
= 4!2! (4-2)!=6

Number of selection of 3 consonants taking from 5 consonants
= ⁵C₃
= 5!3! (5-3)! = 10

∴ Total number of words that can be formed is 6×10×5!= 7200

Page. 16.46 (Multiple Choice Questions)

Q3.

Answer :

(a) 9 !(2 !)3
The word COMMITTEE consists of 9 letters including two Ts, two Ms and two Es.
Number of words that can be formed = Number of arrangements of 9 things taking all at a time, of which 2 are similar of the first kind, 2 are similar of the second kind and 2 are similar of the third kind = 9!2!2!2!=9!(2!)3

Q4.

Answer :

(b) 360
10 lakhs consists of seven digits.
Number of arrangements of seven numbers of which 2 are similar of first kind, 3 are similar of second kind = 7!2!3!
But, these numbers also include the numbers in which the first digit has been considered as 0. This will result in a number less than 10 lakhs. Thus, we need to subtract all those numbers.
Numbers in which the first digit is fixed as 0 = Number of arrangements of the remaining 6 digits = 6!2!3!
Total numbers greater than 10 lakhs that can be formed using the given digits = 7!2!3!-6!2!3!
= 420-60
= 360

Q5.

Answer :

(b) 1956
Number of permutations of six signals taking 1 at a time = 6P1
Number of permutations of six signals taking 2 at a time = 6P2
Number of permutations of six signals taking 3 at a time = 6P3
Number of permutations of six signals taking 4 at a time = 6P4
Number of permutations of six signals taking 5 at a time = 6P5
Number of permutations of six signals taking all at a time = 6P6

∴ Total number of signals =6!5!+6!4!+6!3!+6!2!+6!1!+6!

= 6 + 30 + 120 + 360 + 720 + 720 = 1956

Q6.

Answer :

(b) 240
Total number of words that can be formed of the letters of the word BHARAT = 6!2!
= 360
Number of words in which the letters B and H are always together = 2× 5!2!
= 120
∴ Number of words in which the letters B and H are never together = 360-120
= 240

Q7.

Answer :

(a) 12
All S’s can be placed either at even places or at odd places, i.e. in 2 ways.
The remaining letters can be placed at the remaining places in 3!, i.e. in 6 ways.
∴ Total number of ways = 6×2 = 12

Q8.

Answer :

(b) 60
There are 4 cases where E precedes I i.e.
Case 1: When E and I are together, which are possible in 4 ways whereas other 3 letters are arranged in 3!,
So, the number of arrangements=4×3!=24
Case 2: When E and I have 1 letter in between, which are possible in 3 ways whereas other 3 letters are arranged in 3!,
So,the number of arrangements=3×3!=18
Case 3: When E and I have 2 letters in between, which are possible in 2 ways whereas other 3 letters are arranged in 3!,
So,the number of arrangements=2×3!=12
Case 4: When E and I have 3 letters in between, which are possible in 1 way whereas other 3 letters are arranged in 3!,
So,the number of arrangements=1×3!=6
Thus, total number of arrangements=24+18+12+6=60

Q9.

Answer :

(a) 360
The word CONSTANT consists of two vowels that are placed at the 2nd and 6th position, and six consonants.
The two vowels can be arranged at their respective places, i.e. 2nd and 6th place, in 2! ways.
The remaining 6 consonants can be arranged at their respective places in 6!2!2!ways.
∴ Total number of arrangements = 2!×6!2!2! = 360

Q10.

Answer :

(a) 120
Total number of arrangements of the letters of the word CHEESE = Number of arrangements of 6 things taken all at a time, of which 3 are of one kind = 6!3! = 120

Q11.

Answer :

(a) 24
In order to make a number divisible by 4, its last two digits must be divisible by 4, which in this case can be 12, 24, 32 or 52.
Since repetition of digits is not allowed, the remaining first two digits can be arranged in 3×2 ways in each case.
∴ Total number of numbers that can be formed = 4×{3×2} = 24

Q12.

Answer :

(a) 324
When arranged alphabetically, the letters of the word KRISNA are A, I, K, N, R and S.

Number of words that will be formed with A as the first letter = Number of arrangements of the remaining 5 letters
= 5!
Number of words that will be formed with I as the first letter = Number of arrangements of the remaining 5 letters
= 5!
∴ The number of words beginning with KA = Number of arrangements of the remaining 4 letters
= 4!
The number of words starting with KI = Number of arrangements of the remaining 4 letters
= 4!
The number of words starting with KN = Number of arrangements of the remaining 4 letters
= 4!

Alphabetically, the next letter will be KR.
Number of words starting with KR followed by A, i.e. KRA = Number of arrangements of the remaining 3 letters
= 3!
Number of words starting with KRI followed by A, i.e. KRIA = Number of arrangements of the remaining 2 letters
= 2!
Number of words starting with KRI followed by N, i.e. KRIN = Number of arrangements of the remaining 2 letters
= 2!
The first word beginning with KRIS is the word KRISAN and the next word is KRISNA.

∴ Rank of the word KRISNA = 5! + 5! + 4! + 4! + 4! + 3! + 2! + 2! + 2
= 324

Q13.

Answer :

(c) 6
According to the question:
ⁿP₄ = 12×ⁿP₂
⇒n!n-4!=12×n!n-2!⇒n-2!n-4!=12⇒n-2n-3 = 4×3⇒n-2 =4⇒n=6

Q14.

Answer :

(a) 4! × 3!
According to the question, 3 men have to be ‘consecutive’ means that they have to be considered as a single man.
But, these 3 men can be arranged among themselves in 3! ways.
And, the remaining 3 men, along with this group, can be arranged among themselves in 4! ways.
∴ Total number of arrangements = 4! × 3!

Q15.

Answer :

(a) 216
A number is divisible by 3 when the sum of the digits of the number is divisible by 3.
Out of the given 6 digits, there are only two groups consisting of 5 digits whose sum is divisible by 3.
1+2+3+4+5 = 15
0+1+2+4+5 = 12
Using the digits 1, 2, 3, 4 and 5, the 5 digit numbers that can be formed = 5!
Similarly, using the digits 0, 1, 2, 4 and 5, the number that can be formed = 5!-4! {since the first digit cannot be 0}
∴ Total numbers that are possible = 5! + 5! – 4! = 240 – 24 = 216

Q16.

Answer :

(a) r !
The product of r consecutive integers is equal to r!, so it will be divisible by r!.

Q17.

Answer :

(b) 6 and 7

^k+5P_k+1 = 11 (k-1)2. ^k+3P_k

⇒k+5!k+5-k-1!=11k-12×k+3!k+3-k!⇒k+5!4!=11k-12×k+3!3!⇒k+5!k+3!=11k-12×4!3!⇒k+5k+4 = 22k-1⇒k2+9k+20 = 22k-22⇒k2-13k+42 = 0⇒k = 6,7

Q18.

Answer :

(a) 2^m+n+p − 1
Each of the object, i.e book, pen or pencil, can either be selected or not selected.
So, each of the object will have two outcomes, i.e selection or rejection.
Number of ways of selecting all the books = 2×2×2×…..m times = 2^m
Number of ways of selecting all the pens = 2ⁿ
Number of ways of selecting all the pencils = 2^p
Thus, by fundamental principle of counting, total number of ways = 2^m×2ⁿ×2^p = 2^m+n+p
But, of all these ways, there is one way in which all the objects are rejected, which is not valid. Hence, it needs to be subtracted.
∴ Total number of valid selections = 2^m+n+p − 1

Page. 16.47 (Multiple Choice Questions)

Q19.

Answer :

(c) 36
The word APURBA is a 6 letter word consisting of 3 vowels that can be arranged in 3 alternate places, in 3!2!ways.
The remaining 3 consonants can be arranged in the remaining 3 places in 3! ways.
∴ Total number of words that can be formed = 3!2!×3! = 18
But this whole arrangement can be set-up in total two ways, i.e either VCVCVC or CVCVCV.
∴ Total number of words = 18×2 = 36

Q20.

Answer :

(a) 60 × 5!
The four people, i.e A, B and the two persons between them are always together. Thus, they can be considered as a single person.
So, along with the remaining 4 persons, there are now total 5 people who need to be arranged. This can be done in 5! ways.
But, the two persons that have to be included between A and B could be selected out of the remaining 6 people in 6P2 ways, which is equal to 30.
For each selection, these two persons standing between A and B can be arranged among themselves in 2 ways.
∴ Total number of arrangements = 5!×30×2 = 60×5!

Q21.

Answer :

(a) 576
There are 3 even places in the 7 letter word ARTICLE.
So, we have to arrange 4 consonants in these 3 places in ⁴P₃ ways.
And the remaining 4 letters can be arranged among themselves in 4! ways.
∴ Total number of ways of arrangement = ⁴P₃ ×4! = 4!×4! = 576
Q22.
Answer :

(c) 2¹² − 1
Each of the bulb has its own switch, i.e each bulb will have two outcomes − it will either glow or not glow.
Thus, each of the 12 bulbs will have 2 outcomes.
∴ Total number of ways to illuminate the room = 2¹²

Here, we have also considered the way in which all the bulbs are switched-off. However, this is not required as we need to find out only the number of ways of illuminating the room.
Hence, we subtract that one way from the total number of ways.
= 2¹² − 1

Q23.

Answer :

(a) 72

When we make words after selecting letters of the word BHARAT, it could consist of a single A, two As or no A.

Case-I: A is not selected for the three letter word.

Number of arrangements of three letters out of B, H, R and T = 4×3×2 = 24

Case-II: One A is selected and the other two letters are selected out of B, H, R or T.
Possible ways of selection: Selecting two letters out of B, H, R or T can be done in P24 =12 ways.
Now, in each of these 12 ways, these two letters can be placed at any of the three places in the three letter word in 3 ways.

∴ Total number of words that can be formed = 12×3 = 36

Case-III: Two A’s and a letter from B, H, R or T are selected.

Possible ways of arrangement:
Number of ways of selecting a letter from B, H, R or T = 4

And now this letter can be placed in any one of the three places in the three letter word other than the two A’s in 3 ways.

∴ Total number of words having 2 A’s = 4×3 = 12

Hence, total number of words that can be formed = 24 + 36 + 12 = 72

Q24.

Answer :

(d) 69760
Total number of five digit numbers (since there is no restriction of the number 0XXXX) = 10×10×10×10×10 = 100000
These numbers also include the numbers where the digits are not being repeated. So, we need to subtract all such numbers.
Number of 5 digit numbers that can be formed without any repetition of digits = 10×9×8×7×6 = 30240
∴ Number of five-digit telephone numbers having at least one of their digits repeated = {Total number of 5 digit numbers} – {Number of numbers that do not have any digit repeated} = 100000 – 30240 = 69760

Q25.

Answer :

(d) 144
The word ARTICLE consists of 3 vowels that have to be arranged in the three even places. This can be done in 3! ways.
And, the remaining 4 consonants can be arranged among themselves in 4! ways.
∴ Total number of ways = 3!×4! = 144

Page. 17.8 Ex 17.1

Q1.

Answer :

(i) We have,
14C3= 143 × 132× 121×11C0 [∵ nCr = nr n-1Cr-1]
⇒14C3 = 364 [∵ nC0 = 1]

(ii) We have,
12C10=12C2 [∵ nCr = nCn-r]

⇒ 12C10=12C2 = 122×111×10C0 [∵ nCr = nr n-1Cr-1]

⇒12C10 = 122×111× 1 [∵ nC0 = 1]

⇒12C10 = 66

(iii) We have,

35C35 = 35C0 [∵ nCr = nCn-r]

⇒35C35 =1 [∵ nC0 = 1]

(iv) We have,

ⁿ⁺¹C_n = ⁿ⁺¹C1 [∵ nCr = nCn-r]
⇒n+1Cn= n+1C1= n+11× nC0 [∵ nCr = nr n-1Cr-1]
⇒ ⁿ⁺¹C_n = n+1 [∵ nC0 = 1]

(v) We have,

∑r=15 5Cr=5C1+5C2+5C3+5C4+5C5

⇒∑r=15 5Cr=5C1+5C3+5C3+5C1+5C0 [∵ nCr = nCn-r]

⇒ ∑r=155Cr = 2×51×4C0 +2×53×42×31×2C0 + 5C0 [∵ nCr = nr n-1Cr-1]

⇒∑r=155Cr = 10 + 20 + 1 = 31. [∵ nC0 = 1]

Q2.

Answer :

We have,
ⁿC₁₂ = ⁿC₅
⇒ n = 12 + 5 = 17 [∵ nCx = nCy ⇒ x = y or, n = x+y]

Q3.

Answer :

We have,
nC4= nC6
⇒ n = 6+4 = 10 [∵ nCx = nCy ⇒ x = y or, n = x+y]

Now, 12C10= 12C2 [∵ nCr = nCn-r]
⇒12C10 =12C2 = 122×111×10C0 [∵ nCr = nr n-1Cr-1]
⇒ 12C10 = 66 [∵ nC0 = 1]

Q4.

Answer :

Given: ⁿC₁₀ = ⁿC₁₂
We have,
nC10 = nC12
⇒ n = 12+10 = 22 [∵ nCx = nCy ⇒ x = y or, n = x+y]

Now, 23C22 = 23C1 [∵ nCr = nCn-r]
⇒ 23C22 =23C1 = 231×22C0 [∵ nCr = nr n-1Cr-1]
⇒23C22 = 23 [∵ nC0 = 1]

Q5.

Answer :

Given:
²⁴C_x = ²⁴C2_2x+3
We have,
24 = x + 2x + 3 [∵ nCx = nCy ⇒ x = y or, n = x+y]
⇒24 = 3x + 3⇒3x = 21⇒x = 7

Q6.

Answer :

Given:
¹⁸C_x = ¹⁸C_x+2

By using nCx = nCy ⇒ x = y or n = x+y we get,
18 = x + x + 2⇒ 2x = 16⇒ x = 8

Q7.

Answer :

Given:
¹⁵C_3r = ¹⁵C_r+3
15 = 3r + r + 3. [∵ nCx = nCy ⇒ x = y or, n = x+y]
⇒ 15 = 4r + 3⇒ 4r = 12⇒ r = 3

Q8.

Answer :

Given:
⁸C_r − ⁷C₃ = ⁷C₂
We have,
8Cr= 7C2 + 7C3
⇒ 8Cr=8C3 [∵ nCr + nCr-1 = n+1Cr ; r≤n]
⇒r = 3 [∵ nCx = nCy ⇒ x = y or, n = x+y]

And r+3=8⇒r=5

Q9.

Answer :

Given:
¹⁵C_r : ¹⁵C_r-1 = 11 : 5

We have,
15Cr15Cr-1 = 115

⇒ 15-r+1r = 115 [∵ nCrnCr-1 = n-r+1r]

⇒ 75 – 5r + 5 = 11r⇒ 16r = 80⇒ r = 5

Q10.

Answer :

We have, ⁿ⁺²C₈ : ^n-2P₄ = 57 : 16

⇒ n+2C8n-2P4 = 5716⇒ (n+2)! 8! (n-6)!×(n-6)!(n-2)! = 5716⇒(n+2) (n+1) n (n-1) (n-2)!8!×1(n-2)! = 5716⇒ (n+2) (n+1) n (n-1) = 5716×8! = 19×316×8×7×6×5×4×3×2×1⇒(n+2) (n+1) n (n-1) = 143640⇒ (n-1) n (n+1) (n+2) = 19×3×7×6×5×4×3⇒(n-1) n (n+1) (n+2) = 19×(3×7)×(6×3)×(4×5)⇒(n-1) n (n+1) (n+2) = 18×19×20×21⇒ n-1 = 18⇒ n = 19

Page. 17.9 Ex 17.1

Q11.

Answer :

We have, ²⁸C_2r : ²⁴C_2r-4 = 225 : 11
⇒28C2r24C2r-4 = 22511⇒ 28!2r! (28-2r)!× (2r-4)! (28-2r)!24! = 22511⇒ 28×27×26×252r (2r-1) (2r-2) (2r-3) = 22511⇒ 2r (2r-1) (2r-2) (2r-3) = 28×27×26×25×11225⇒ 2r (2r-1) (2r-2) (2r-3) = 28×3×26×11⇒ 2r (2r-1) (2r-2) (2r-3) = 4×7×3×13×2×11⇒ 2r (2r-1) (2r-2) (2r-3) = (2×7)×13×(3×4)×11⇒ 2r (2r-1) (2r-2) (2r-3) = 14×13×12×11⇒ 2r = 14⇒r = 7

Q12.

Answer :

4nC2n2nCn=1.3.5…4n-11.3.5…2n-12LHS=4nC2n2nCn =4n!2n!2n!×n!n!2n! =4n×4n-1×4n-2×4n-3……………..3×2×1 ×n!22n×2n-1×2n-2…….3×2××122n! =1×3×5……..4n-12×4×6……………4n×n!21×3×5×………2n-122×4×6×…….2n2×2n! =1×3×5……..4n-1×22n×1×2×3……….2nn!21×3×5×………2n-12×22n1×2×3×…….n22n! =1×3×5……..4n-12n!n!21×3×5×………2n-12n!22n! =1×3×5……..4n-11×3×5×………2n-12 =RHSHence, proved.

Q13.

Answer :

Given: 2nC3:nC2 = 44:3
2nC3nC2 = 443⇒2n!3! (2n-3)!× 2! (n-2)!n! = 443⇒ 2n (2n-1) (2n-2)3 n (n-1) = 443⇒ (2n-1) (2n-2) = 22 (n-1)⇒ 4n2 – 6n + 2 = 22n – 22⇒ 4n2 – 28n + 24 = 0⇒ n2 – 7n + 6 = 0⇒ n2 – 6n – n + 6 = 0⇒ n (n-6) -1(n-6) = 0⇒ (n-1) (n-6) = 0
⇒n =1 or, n= 6

Now, n=1 ⇒ 2C3:2C2 = 44:3
But, this is not possible.
∴ n = 6

Q14.

Answer :

Given:
16Cr = 16Cr+2

16 = r + r + 2 [∵ Property 5: nCx = nCy ⇒ x = y or x + y = n]
⇒2r + 2 = 16⇒ 2r = 14⇒ r = 7

Now,rC4 = 7C4
⇒ 7C4 = 7C3 [∵ nCr =nCn-r]
⇒7C4 = 7C3 = 73×62×51×4C0 [∵ nCr = nr . n-1Cr-1 ]
⇒7C4= 35 [∵ nC0 = 1]

Q15.

Answer :

Given:
20C5 + ∑r=25 25-rC4

20C5 + ∑r=25 25-rC4= 20C5+23C4 + 22C4 + 21C4 + 20C4=20C4 + 20C5 +21C4+22C4+23C4

=21C5 +21C4 + 22C4+23C4 [∵ nCr-1 +nCr =n+1Cr]
=21C4 + 21C5 +22C4+23C4
=22C5+22C4+23C4 [∵nCr-1 +nCr =n+1Cr]
=23C5+22C4+23C4
=23C5+23C4=24C5 [∵ nCr-1 +nCr =n+1Cr]

=25!19! 5! = 24×23×22×21×205×4×3×2×1 = 42504

Q16.

Answer :

Let 2n negative integers be -r,-r-1, -r-2,….,…, -r-2n+1.

Then, product = -12nrr+1r+2,….,…r+2n-1
= r-1! rr+1 r+2 ……r+2n-1r-1!= r+2n-1!r-1!= r+2n-1!r-1!2n!×2n!= r+2n-1C2n×2n!
This is divisible by 2n!.

Q17.

Answer :

LHS = 2nCn+ 2nCn-1 =2n!n! n! + 2n!n-1! 2n – n +1! = 2n!n! n! + 2n!n-1! n+1! = 2n!n n-1! n! + 2n!n-1! n+1n! = 2n!n! n-1! 1n + 1n+1 = 2n!n! n-1! 2n+1n n+1 = 2n+1!n! n+1!

RHS = 12 2n+2Cn+1 = 12 2n+2!n+1! 2n + 2 – n-1! = 12 2n+2!n+1! n+1! = 12 2n+2 2n+1!n+1 n! n+1! = 12 2n+1 2n+1!n+1 n! n+1! = 2n+1!n! n+1!

∴ LHS = RHS

Q18.

Answer :

Since ⁿC₄ , ⁿC₅ and ⁿC₆ are in AP.

∴ 2. ⁿC₅ = ⁿC₄ + ⁿC₆
⇒2×n!5!n-5!=n!4!n-4!+n!6!n-6!⇒25×4!n-5n-6!=14!n-5n-4n-6!+16×5×4!n-6!⇒25n-5=1n-5n-4+130⇒25n-5-1n-5n-4=130⇒2n-8-55n-5n-4=130⇒2n-13n-5n-4=16⇒12n-78=n2-9n+20⇒n2-21n+98=0⇒n-7n-14=0∴ n=7 and 14

Q19.

Answer :

αC2 = α2×(α-1)1×αC0 [∵ nCr = nr. n-1Cr-1]

= 12 α (α-1) [∵ nC0 = 1]

= 12 mC2 mC2 -1=12 m!2! m-2! m!2! m-2! -1= 12 m m-12 m m-12 -1=12 mm-12 m m-1 – 22 = 18 m m-1 m m-1 -2=18 m2 m-12 – 2m m -1= 18 m2 m2 + 1 – 2m -2m2 + 2m= 18 m4 + m2 – 2m3 – 2m2 + 2m= 18 m4 -2m3 – m2 + 2m= 18 m2 – 2m m2 -1= 18 m m-2 m-1 m+1= 18 m+1 m m-1 m-2

Q20.

Answer :

(a)
nCrnCr-1 = n-r+1r

LHS = nCrnCr-1 = n!r! n-r!×r-1! n-r+1!n! = n-r+1 n-r! r-1!r r-1! n-r! = n-r+1r = RHS

∴ LHS = RHS

(b)
LHS= n. n-1Cr-1 = n n-1!r-1! n-1-r+1! = n!r-1! n-r!RHS=n-r+1 nCr = n-r+1 n!r-1! n-r+1! =n-r+1n!r-1! n-r+1n-r! = n!r-1! n-r!

∴ LHS = RHS

(c)
nCrn-1Cr-1 = nr

LHS = nCrn-1Cr-1 = n!r! n-r!×r-1! n-1-r+1!n-1! =n n-1!r r-1! n-r! × r-1! n-r!n-1! = nr = RHS

∴ LHS = RHS

(d)
nCr+ 2. nCr-1 +nCr-2 = n+2Cr

LHS =nCr +2.nCr-1 +nCr-2 = nCr +nCr-1 + nCr-1 +nCr-2
= n+1Cr + n+1Cr-1 [∵ nCr + nCr-1 = n+1Cr]
=n+2Cr [∵ nCr + nCr-1 = n+1Cr]
= RHS
∴ LHS = RHS

Page. 17.16 Ex 17.2

Q1.

Answer :

Required number of ways = 15C11

Now,15C11 =15C4
= 154×143×132×121×11C0

= 1365

Q2.

Answer :

Clearly, out of the 25 boys and 10 girls, 5 boys and 3 girls will be chosen.

Then, different boat parties of 8 = 25C5×10C3

=25!5! 20!×10!3! 7!= 25×24×23×22×215×4×3×2×1×10×9×83×2×1= 6375600

Q3.

Answer :

We are given that 2 courses are compulsory out of the 9 available courses,
Thus, a student can choose 3 courses out of the remaining 7 courses.
Number of ways = 7C3 = 7!3! 4! = 7×6×5×4!3×2×1×4! = 35

Q4.

Answer :

Number of ways in which 11 players can be selected out of 16 = 16C11 = 16!11! 5! = 16×15×14×13×125×4×3×2×1 = 4368
(i) If 2 particular players are included, it would mean that out of 14 players, 9 players are selected.
Required number of ways = 14C9 = 14!9! 5! = 14×13×12×11×105×4×3×2×1 = 2002

(ii) If 2 particular players are excluded, it would mean that out of 14 players, 11 players are selected.
Required number of ways = 14C11= 14!11! 3! = 14×13×123×2×1 = 364

Page. 17.17 Ex 17.2

Q5.

Answer :

Clearly, 2 professors and 3 students are selected out of 10 professors and 20 students, respectively.
Required number of ways = 10C2×20C3 = 102×91×203×192×181 = 51300

(i) If a particular professor is included, it means that 1 professor is selected out of the remaining 9 professors.
Required number of ways = 20C3×9C1 = 203×192×181×91 = 10260

(ii) If a particular student is included, it means that 2 students are selected out of the remaining 19 students.
Required number of ways = 19C2×10C2 = 192×181×102×91 = 7695

(iii) If a particular student is excluded, it means that 3 students are selected out of the remaining 19 students.
Required number of ways = 19C3×10C2 = 193×182×171×102×91 = 43605

Q6.

Answer :

Required number of ways of getting different products = 4C2+4C3 +4C4 = 6 + 4 + 1 = 11

Q7.

Answer :

Two girls who won the prizes last year are to be included in every selection.
So, we have to select 8 students out of 12 boys and 8 girls, choosing at least 4 boys and 2 girls.
Number of ways in which it can be done = 12C6×8C2 + 12C5×8C3 + 12C4×8C4 = 25872 + 44352 + 34650 = 104874

Q8.

Answer :

(i) Required ways of selecting 4 books from 10 books without any restriction = 10C4 = 104×93×82×7 = 210

(ii) Two particular books are selected from 10 books. So, 2 books need to be selected from 8 books.
Required number of ways if 2 particular books are always selected = 8C2 = 82×71 = 28

(iii) Two particulars books are never selected from 10 books. So, 4 books need to be selected from 8 books.
Required number of ways if two particular books are never selected = 8C4 = 84×73×62×51 = 70

Q9.

Answer :

(i) From 4 officers and 8 jawans, 6 need to be chosen. Out of them, 1 is an officer.
Required number of ways = 4C1 × 8C5 = 4 × 8!5! 3! = 4×8×7×6×5!5! ×6 = 224

(ii) From 4 officers and 8 jawans, 6 need to be chosen and at least one of them is an officer.
Required number of ways = Total number of ways – Number of ways in which no officer is selected
=12C6 -8C6 = 12!6! 6! – 8!6! 2! = 12×11×10×9×8×76×5×4×3×2×1 – 8×72 = 924-28=896

Q10.

Answer :

A sports team of 11 students is to be constituted, choosing at least 5 students of class XI and at least 5 from class XII.
Required number of ways = 20C5×20C6 + 20C6×20C5 = 2 ×20C5×20C6 = 2 20C6 × 20C5

Q11.

Answer :

The various possibilities for answering the 10 questions are given below:
(i) 4 from part A and 6 from part B.
(ii) 5 from part A and 5 from part B.
(iii) 6 from part A and 4 from part B.
∴ Required number of ways = 6C4×7C6 + 6C5×7C5 + 6C6×7C4

=6!4! 2! × 7+ 6 × 7!5! 2! + 1 × 7!4! 3! =105+126+35=266

Q12.

Answer :

A student has to answer 4 question out of 5 questions.
Since questions 1 and 2 are compulsory, he/she will have to answer 2 question from the remaining 3.
∴ Required number of ways =3C2 = 3

Q13.

Answer :

Required ways = 6C5×6C2 +6C4×6C3 + 6C3×6C4 + 6C2×6C5
=26C5×6C2+6C4×6C3 =290+300=2390=780

Q14.

Answer :

Number of straight lines formed joining the 10 points, taking 2 points at a time = 10C2 = 102×91 = 45
Number of straight lines formed joining the 4 points, taking 2 points at a time = 4C2 = 42× 31 = 6
But, when 4 collinear points are joined pair wise, they give only one line.
∴ Required number of straight lines = 45- 6 +1 = 40

Q15.

Answer :

A polygon of n sides has n vertices. By joining any two vertices we obtain either a side or a diagonal.
∴ Number of ways of selecting 2 out of 9 =nC2=nn-12
Out of these lines, n lines are the sides of the polygon.

∴ Number of diagonals = nn-12-n=nn-32

(i) In a hexagon, there are 6 sides.
∴ Number of diagonals = 66-32=9

(ii) There are 16 sides.
∴ Number of diagonals = 1616-32=104

Q16.

Answer :

Out of 12 points, 5 points are collinear and 3 points are required to form a triangle.
Required ways =12C3 -5C3 = 123×112×101 – 53×42×31 = 220 – 10 = 210

Q17.

Answer :

5 persons are to be selected out of 6 men and 4 women. At least, one woman has to be selected in all cases.

Required number of ways =4C1×6C4 + 4C2×6C3 + 4C3×6C2 +4C4×6C1= 60 + 120 + 60 + 6 = 246

Q18.

Answer :

52 families have at most 2 children, while 35 families have more than 2 children.
The selection of 20 families of which at least 18 families must have 2 children can be made in the ways given below.
(i) 18 families out of 52 and 2 families out of 35
(ii) 19 families out of 52 and 1 family out of 35
(iii) 20 families out of 52
∴ Required ways = 52C18 × 35C2 + 52C19 × 35C1 + 52C20 × 35C0

Q19.

Answer :

Each set of parallel lines consists of m+2 lines.
Each parallelogram is formed by choosing two lines from the first set and two straight lines from the second set.
∴ Total number of parallelograms = m+2C2 × m+2C2 = m+2C22

Q20.

Answer :

A decagon has 10 sides.
(i) Number of diagonals = n n-32 = 10 10-32 = 35
(ii) Number of triangles (i.e. 3 sides are to be selected) = 10C3 = 103×92×81 = 120

Q21.

Answer :

(i) Number of straight lines formed joining the 18 points, taking 2 points at a time = 18C2 = 182×171 = 153
Number of straight lines formed joining the 5 points, taking 2 points at a time = 5C2 = 52× 41 = 10
But, when 5 collinear points are joined pair wise, they give only one line.
∴ Required number of straight lines = 153 – 10 +1 = 144

(ii) Number of triangles formed joining the 18 points, taking 3 points at a time = 18C3 = 183×172×161 = 816
Number of straight lines formed joining the 5 points, taking 3 points at a time = 5C3 = 53× 42×31 = 10
∴ Required number of triangles = 816 – 10= 806

Page. 17.18 Ex 17.2

Q22.

Answer :

There are 4 kings in the deck of cards.
So, we are left with 48 cards out of 52.
∴ Required combination = 48C1×4C4 + 48C2×4C3 + 48C3×4C2 + 48C4×4C1
= 48 + 4512 + 103776 + 778320= 886656

Q23.

Answer :

6 people are to be selected from 8.

There are two case.
(i) When A is selected, then B must be chosen.
∴ Number of ways= 6C4=15

(ii) When A is not chosen:
Number of ways=7C6=7

∴ Total number of ways = 15 + 7 = 22

Q24.

Answer :

3 boys and 3 girls are to be selected from 5 boys and 4 girls.
∴ Required ways = 5C3×4C3 = 53×42×31×43×32×21 =40

Q25.

Answer :

Required number of ways = 6C3×5C3×5C3 = 6!3! 3!× 5!3! 2!×5!3! 2! = 2000

Q26.

Answer :

There are total 4 aces in the deck of 52 cards. So, we are left with 48 cards.
∴ Required ways = 4C1×48C4= 41×484×473×462×451 = 778320

Q27.

Answer :

Out of 17 players, 11 need to be selected. There are 5 bowlers, of which four must be selected in the team. So, we have to choose 7 players from the remaining 12 players.
Required number of ways = 5C4×12C7 = 5× 12!7! 5! = 5×792 = 3960

Q28.

Answer :

2 black and 3 red balls are to be selected from 5 black and 6 red balls.
Required number of ways = 5C2×6C3 = 52×41×63×52×41 = 200

Q29.

Answer :

2 courses are compulsory out of the 9 available courses. There are 7 more courses.
So, we need to choose 3 courses out of 7 courses.
∴ Required number of ways = 7C3 = 73×62×51 = 35

Q30.

Answer :

A committee of 7 has to be formed from 9 boys and 4 girls.

(i) When the committee consists of exactly 3 girls:
Required number of ways = 4C3×9C4 = 43×32×21×94×83×72×61 = 504

(ii) When the committee consists of at least 3 girls:
Required number of ways = 4C3×9C4 + 4C4×9C3 =504 + 84 = 588

(iii) When the committee consists of at most 3 girls:

Required number of ways = 4C0×9C7 + 4C1×9C6 + 4C2×9C5 +4C3×9C4 = 36 + 336 + 756 + 504 = 1632

Q31.

Answer :

A question paper consists of 12 questions divided into 2 parts, one with 5 and the other with 7 questions.
A student has to attempt 8 questions out of the 12 questions by selecting at least 3 from each part.
∴ Required number of ways =5C3×7C5 + 5C4×7C4+5C5×7C3 = 210 + 175 + 35 = 420

Q32.

Answer :

A group consists of 4 girls and 7 boys. Out of them, 5 are to be selected to form a team.
(i) If the team has no girls, then the number of ways of selecting 5 members =7C5 = 7!5! 2! = 7×62 = 21

(ii) If the team has at least 1 boy and 1 girl, then the number of ways of selecting 5 members
= 4C1×7C4+4C2×7C3 + 4C3×7C2 +4C4×7C1 = 140 + 210+ 84 + 7 = 441

(iii) If the team has at least 3 girls, then the number of ways of selecting 5 members
=4C3×7C2 +4C4×7C1 = 84 + 7 = 91

Q33.

Answer :

A committee of 3 people is to be constituted from a group of 2 men and 3 women.
∴ Number of ways =2C0×3C3+2C1×3C2 + 2C2×3C1=1+ 2 × 3 + 3×1=10

Number of committees consisting of 1 man and 2 women =2C1×3C2 = 6

Page. 17.24 Ex 17.3

Q1.

Answer :

2 out of 5 vowels and 3 out of 17 consonants can be chosen in5C2× 17C3 ways.
Thus, there are5C2× 17C3 groups, each containing 2 vowels and 3 consonants.
Each group contains 5 letters, which can be arranged in 5! ways.
∴ Required number of words = 5C2× 17C35! = 6800 ×120 = 816000

Q2.

Answer :

A tea party is arranged for 16 people along two sides of a long table with 8 chairs on each side.
4 people wish to sit on side A (say) and two on side B (say).
Now, 10 people are left, out of which 4 people can be selected for side A in 10C4 ways.
And, from the remaining people, 6 people can be selected for side B in 6C6 ways.
∴ Number of selections =10C4× 6C6
Now, 8 people on each side can be arranged in 8! ways.
∴ Total number ways in which the people can be seated =10C4× 6C6× 8! × 8! = 10C4×8!2

Q3.

Answer :

A businessman hosts a dinner for 21 guests.
15 people can be accommodated at one table in 21C15 ways. They can arrange themselves in 15-1!=14! ways.
The remaining 6 people can be accommodated at another table in 6C6 ways. They can arrange themselves in 6-1!=5! ways.
∴ Total number of ways =21C15×6C6×14!×5!=21C15×14!×5!

Q4.

Answer :

There are 11 letters in the word EXAMINATION, namely AA, NN, II, E, X, M, T and O.
The four-letter word may consist of
(i) 2 alike letters of one kind and 2 alike letters of the second kind
(ii) 2 alike letters and 2 distinct letters
(iii) all different letters
Now, we shall discuss the three cases one by one.

(i) 2 alike letters of one kind and 2 alike letters of the second kind:
There are three sets of 2 alike letters, namely AA, NN and II.
Out of these three sets, two can be selected in 3C2 ways.
So, there are 3C2 groups, each containing 4 letters out of which two are alike letters of one kind and two 2 are alike letters of the second kind.
Now, 4 letters in each group can be arranged in 4!2! 2! ways.
∴ Total number of words that consists of 2 alike letters of one kind and 2 alike letters of the second kind =3C2×4!2! 2! = 3×6 = 18

(ii) 2 alike and 2 different letters:
Out of three sets of two alike letters, one set can be chosen in 3C1 ways.
Now, from the remaining 7 letters, 2 letters can be chosen in 7C2 ways.
Thus, 2 alike letters and 2 distinct letters can be chosen in 3C1×7C2 ways.
So, there are 3C1×7C2 groups of 4 letters each.
Now, the letters in each group can be arranged in 4!2! ways.
∴ Total number of words consisting of 2 alike and 2 distinct letters = 3C1× 7C2 ×4!2! = 756
(iii) All different letters:
There are 8 different letters, namely A, N, I, E, X, M, T and O. Out of them, 4 can be selected in 8C4 ways.
So, there are 8C4 groups of 4 letters each. The letters in each group can be arranged in 4! ways.
∴ Total number of four-letter words in which all the letters are distinct =8C4× 4! = 1680

∴ Total number of four-letter words = 18 + 756 + 1680 = 2454

Q5.

Answer :

There are 10 letters in the word PROPORTION, namely OOO, PP, RR, I, T and N.
(a) The four-letter word may consists of
(i) 3 alike letters and 1 distinct letter
(ii) 2 alike letters of one kind and 2 alike letters of the second kind
(iii) 2 alike letters and 2 distinct letters
(iv) all distinct letters

Now, we shall discuss these four cases one by one.
(i) 3 alike letters and 1 distinct letter:
There is one set of three alike letters, OOO, which can be selected in one way.
Out of the 5 different letters, P, R, I, T and N, one can be selected in5C1 = 5 ways.
(ii) There are 3 sets of two alike letters, which can be selected in 3C2 = 3 ways.
(iii) There are three sets of two alike letters, which can be selected in 3C1 ways.
Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways.
Thus, 2 alike letters and 2 different letters can be selected in 3C1×5C2 = 30 ways.
(iv) There are 6 different letters.
Number of ways of selecting 4 letters = 6C4 = 15

∴ Total number of ways = 5+ 3 + 30 + 15 = 53

(b) The four-letter word may consists of
(i) 3 alike letters and 1 distinct letter
(ii) 2 alike letters of one kind and 2 alike letters of the second kind
(iii) 2 alike letters and 2 distinct letters
(iv) all distinct letters

Now, we shall discuss these four cases one by one.
(i) 3 alike letters and 1 distinct letter:
There is one set of three alike letters, OOO, which can be selected in one way.
Out of the 5 different letters, P, R, I, T and N, one can be selected in5C1 ways.
These four letters can be arranged in 4!3! 1! ways.
∴ Total number of ways =5C1×4!3! 1!=20
(ii) There are 3 sets of two alike letters, which can be selected in 3C2 ways.
Now, the letters of each group can be arranged in 4!2! 2! ways.
∴ Total number of ways =3C2×4!2! 2!=18
(iii) There are three sets of two alike letters, which can be selected in 3C1 ways.
Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways.
Thus, 2 alike letters and 2 different letters can be selected in 3C1×5C2 = 30 ways.
Now, the letters of each group can be arranged in 4!2! ways.
∴ Total number of ways = 30×4!2!=360
(iv) There are 6 different letters.
So, the number of ways of selecting 4 letters is 6C4 = 15 and these letters can be arranged in 4! ways.
∴ Total number of ways = 15 × 4! = 360

∴ Total number of ways = 20 + 18 + 360 + 360 = 758

Q6.

Answer :

There are 9 letters in the word MORADABAD, namely AAA, DD, M, R, B and O.
The four-letter word may consists of
(i) 3 alike letters and 1 distinct letter
(ii) 2 alike letters of one kind and 2 alike letters of the other kind
(iii) 2 alike letters and 2 distinct letters
(iv) all different letters

(i) 3 alike letters and 1 distinct letter:
There is one set of three alike letters, AAA, which can be selected in one way.
Out of the 5 different letters D, M, R, B and O, one can be selected in5C1 ways.
These four letters can be arranged in 4!3! 1! ways.
∴ Total number of ways =5C1×4!3! 1!=20

(ii) There are two sets of two alike letters, which can be selected in 2C2 ways.
Now, the letters of each group can be arranged in 4!2! 2! ways.
∴ Total number of ways =2C2×4!2! 2!=6

(iii) There is only one set of two alike letters, which can be selected in 2C1 ways.
Now, from the remaining 5 letters, 2 letters can be chosen in 5C2 ways.
Thus, 2 alike letters and 2 different letters can be selected in 2C1×5C2 = 20 ways.
Now, the letters of each group can be arranged in 4!2! ways.
∴ Total number of ways = 20×4!2!=240

(iv) There are 6 different letters A, D, M,B, O and R.
So, the number of ways of selecting 4 letters is 6C4, i.e. 15, and these letters can be arranged in 4! ways.
∴ Total number of ways = 15 × 4! = 360
∴ Total number of ways = 20 + 6 + 240 + 360 = 626

Q7.

Answer :

There are 4 vowels and 4 consonants in the word INVOLUTE.
Out of these, 3 vowels and 2 consonants can be chosen in 4C3×4C2 ways.
The 5 letters that have been selected can be arranged in 5! ways.
∴ Required number of words = 4C3× 4C2 × 5! = 4×6×120 = 2880

Q8.

Answer :

We have n different things.
We are to select r things at a time such that two specified things occur together.
Remaining things = n -2
Out of the remaining (n -2) things, we can select (r -2) things in n -2Cr -2 ways.
Consider the two things as one and mix them with (r -2) things.
Now, we have (r -1) things that can be arranged in (r -1)! ways.
But, two things can be put together in 2! ways.

∴ Required number of ways = n-2Cr-2×r-1!×2! =2r-1n-2Cr-2×r-2! =2r-1n-2Pr-2

Q9.

Answer :

There are six letters in the word MONDAY.

(i) 4 letters are used at a time:
Four letters can be chosen out of six letters in 6C4 ways.
So, there are ⁶C₄ groups containing four letters that can be arranged in 4! ways.
∴ Number of ways =⁶C₄× 4! = 6!4! 2!× 4! = 6!2! = 360

(ii) All the letters are used at a time:
This can be done in ⁶C₆ ways.
So, there are ⁶C₆ groups containing six letters that can be arranged in 6! ways.
∴ Number of ways =⁶C₆ × 6! = 1× 720 = 720

(iii) All the letters are used, but the first letter is a vowel:
There are two vowels, namely A and O, in the word MONDAY.
For the first letter, out of the two vowels, one vowel can be chosen in 2C1 ways.
The remaining five letters can be chosen in 5C5 ways.
So, the letters in 5C5 group can be arranged in 5! ways.
∴ Number of ways =²C₁×⁵C₅× 5! = 2×1×5! = 240

Page. 17.26 (Multiple Choice Questions)

Q2.

Answer :

(b) 56
r + r + 4 = 20 [∵nCx=nCy ⇒n=x+y or x=y]
⇒2r + 4 = 20⇒ 2r = 16⇒ r = 8

Now,rC3 = 8C3
8C3 = 8!3! 5! = 8×7×63×2×1 = 56

Q3.

Answer :

(c) 3
3r + r + 3 = 15 [∵nCx=nCy ⇒n=x+y or x=y]
⇒ 4r + 3 = 15⇒ 4r = 12⇒ r = 3

Q4.

Answer :

(a) 10
r + 1 + r – 1 = 20 [∵nCx=nCy ⇒n=x+y or x=y]
⇒ 2r = 20⇒ r = 10

Q5.

Answer :

(a) 231
nC12 =nC8
⇒ n = 12 + 8 = 20 [∵nCx=nCy ⇒n=x+y or x=y]

Now,22Cn =22C20

= 222×211
= 231

Q6.

Answer :

(c) 2 m = n (n − 1)

^mC₁ = ⁿC₂
⇒ m!1! m-1! = n!2! n-2!⇒ m m-1!m-1! = n n-1 n-2!2 n-2!⇒ 2m = n n-1

Q7.

Answer :

(a) 20

n = 12 + 8 = 20 [∵nCx=nCy ⇒n=x+y or x=y]

Q8.

Answer :

(d) r + 1

nCr + nCr+1 = n+1Cx [Given]

We have:
nCr +nCr+1 = n+1Cx [∵nCr +nCr-1 = n+1Cr]
⇒ n+1Cr+1 = n+1Cx
⇒r+1= x [∵nCx=nCy ⇒n=x+y or x=y]

Q9.

Answer :

(b) 3

a2-a =2 + 4 [∵nCx=nCy⇒n=x+y or x=y]
⇒ a2 – a – 6 = 0⇒a2 – 3a + 2a – 6 = 0⇒ a a-3 + 2 a-3 = 0⇒ a+2 a-3 = 0
⇒ a=-2 or a=3
But, a=-2 is not possible.
∴ a = 3

Q10.

Answer :

(b) 31

5C1 + 5C2 + 5C3 + 5C4 + 5C5
= 5C1 + 5C2 + 5C2 + 5C1 + 5C5

= 2 × 5C1 + 2 × 5C2 + 5C5= 2 × 5 + 2 × 5!2! 3! + 1 = 10 + 20 + 1 = 31

Q11.

Answer :

(c) 7200
2 out of 4 vowels can be chosen in 4C2 ways and 3 out of 5 consonants can be chosen in 5C3 ways.
Thus, there are C2× 5C43 groups, each containing 2 vowels and 3 consonants.
Each group contains 5 letters that can be arranged in 5! ways.
∴ Required number of words = 4C2× 5C3× 5! = 60 × 120 = 7200

Q13.

Answer :

(a) 60
Three persons can take 5 seats in ⁵C₃ ways. Moreover, 3 persons can sit in 3! ways.
∴ Required number of ways = 5C3× 3! = 10×6 = 60

Q14.

Answer :

(a) 246

Required number of ways=4C1× 6C4 +4C2× 6C3 +4C3× 6C2 +4C4× 6C1 = 60 + 120 + 60 + 6 = 246

Q15.

Answer :

(b) 40
Number of straight lines formed by joining the 10 points if we take 2 points at a time =10C2 = 102×91 = 45
Number of straight lines formed by joining the 4 points if we take 2 points at a time =4C2 = 42× 31 = 6
But, 4 collinear points, when joined in pairs, give only one line.
∴ Required number of straight lines = 45 – 6 +1 = 40

Q16.

Answer :

(b) 78

4 out of 13 players are bowlers.
In other words, 9 players are not bowlers.
A team of 11 is to be selected so as to include at least 2 bowlers.

∴ Number of ways =4C2×9C9 + 4C3×9C8 + 4C4×9C7 = 6 + 36 + 36 = 78

Q17.

Answer :

(b) 150

If set S has n elements, then C n, k is the number of ways of choosing k elements from S.
Thus, the number of subsets of S of all possible values is given by
Cn,0 + Cn,1 + Cn,3 + … + Cn,n = 2n
Comparing the given equation with the above equation:
2n = 256⇒ 2n = 28⇒ n = 8

∴ 2nC2 = 16C2⇒16C2 = 16!2! 14! = 16×152 = 120

Page. 17.27 (Multiple Choice Questions)

Q18.

Answer :

(c) ¹²C₈ − ¹⁰C₆

A host lady can invite 8 out of 12 people in 12C8 ways. Two out of these 12 people do not want to attend the party together.
∴ Number of ways = ¹²C₈ − ¹⁰C₆

Q19.

Answer :

(b) 156
We need at least three points to draw a circle that passes through them.
Now, number of circles formed out of 11 points by taking three points at a time = ¹¹C₃ = 165
Number of circles formed out of 5 points by taking three points at a time = ⁵C₃ = 10
It is given that 5 points lie on one circle.

∴ Required number of circles = 165- 10 + 1 = 156

Q20.

Answer :

(d) 120

Number of committes that can be formed = 6C3 × 4C2 = 6!3! 3! × 4!2! 2! = 6 × 5 × 43 × 2 × 4 × 32 = 120

Q21.

Answer :

(a) 12

r – 6 + 3r + 1 = 43 [∵nCx=nCy ⇒n=x+y or x=y]
⇒ 4r – 5 = 43 ⇒ 4r = 48⇒ r = 12

Q22.

Answer :

(a) 20

An octagon has 8 vertices.
The number of diagonals of a polygon is given by n n-32.
∴ Number of diagonals of an octagon = 8 8-32 = 20

Q23.

Answer :

(b) 2⁸ − 2

C07+C17+C17+C27+C27+C37+C37+C47+C47+C57+C57+C67+C67+C77

=1 + 2 × C17 + 2 × C27 + 2 × C37 + 2 × C47 + 2 × C57 + 2 × C67 + 1

=1 +2 × C17 + 2 × C27 + 2 × C37 + 2 × C37 + 2 × C27 + 2 × C67 + 1

= 2 + 22 C17 + C27 + C37 = 2 + 22 7 + 72 × 6 + 73 × 62 × 5

= 2 + 252 = 254 = 28 – 2

Q24.

Answer :

(c) 264

Among 14 players, 5 are bowlers.
A team of 11 players has to be selected such that at least 4 bowlers are included in the team.
∴ Required number of ways=C45×C79 + C55× C69 = 180 + 84 = 264

Q25.

Answer :

(b) 140

Suppose there are two friends, A and B, who do not attend the party together.
If both of them do not attend the party, then the number of ways of selecting 6 guests = ⁸C₆ = 28
If one of them attends the party, then the number of ways of selecting 6 guests = 2.⁸C₅ = 112
∴ Total number of ways = 112 + 28 = 140

Q26.

Answer :

(c) 5
C3n+1 =2×C2n⇒ n+1!3! n-2! = 2× n!2! n-2!⇒ n+1 n!3×2! n-2! = 2× n!2! n-2!⇒ n+1 = 6⇒ n = 5

Q27.

Answer :

(d) 18
A parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two parallel lines from the set of three parallel lines.
Two parallel lines from the set of four parallel lines can be chosen in ⁴C₂ ways and two parallel lines from the set of 3 parallel lines can be chosen in ³C₂ ways.
∴ Number of parallelograms that can be formed = C24× C23 = 4!2! 2!×3!2! 1! =6×3 = 18