Chapter 7: Permutations and Combinations

According to the question,

Number of mice =18,

Number of groups = 3

Since the groups are equally large,

The number of mice in each group can be = 6 mice

The number of ways of placement of mice =18!

For each group the placement of mice = 6!

Hence, the required number of ways = 18!/(6!6!6!)

= 18!/(6!)³

We know that,

ⁿC_r

According to the question,

Number of white marbles = 6,

Number of red marbles = 5

Total number of marbles = 6 white + 5 red = 11 marbles

(a)If they can be of any colour

Then, any 4 marbles out of 11 can be selected

Therefore, the required number of ways =¹¹C₄

(b) two must be white and two red

Number of ways of choosing two white and two red = ⁶C₂ × ⁵C₂

(c) they must all be of the same colour

Then, four white marbles out of 6 can be selected = ⁶C₄

Or, 4 red marbles out of 5 can be selected = ⁵C₄

Therefore, the required number of ways = ⁶C₄+⁵C₄

We know that,

ⁿC_r

According to the question,

11 players can be selected out of 16 = ¹⁶C₁₁

(i) include 2 particular players = ¹⁴C₉

(ii) exclude 2 particular players = ¹⁴C₁₁

We know that,

ⁿC_r

A team of 11 students can be constituted in the following two ways

(i) 5 students from class XI and 6 students from XII = ²⁰C_5. ²⁰C₆

(ii) 6 students from class XI and 5 students from XII = ²⁰C_6.²⁰C₅

The number ways the team can be constituted = Case(i)+case(ii)

= ²⁰C_5. ²⁰C₆ + ²⁰C_6.²⁰C₅

= 2(²⁰C_5. ²⁰C₆)

We know that,

ⁿC_r

(i) No girls

Total number of ways the team can have no girls = ⁴C_0. ⁷C₅ =21

(ii) at least one boy and one girl

Case(A)1 boy and 4 girls

=⁷C_1. ⁴C₄

=7

Case(B)2 boys and3 girls

=⁷C_2. ⁴C₃

=84

Case(C) 3boys and 2girls

=⁷C_3. ⁴C₂

=210

Case(D)4 boys and 1 girls

=⁷C_4. ⁴C₁

=140

Total number of ways the team can have at least one boy and one girl,

= Case(A) + Case(B) + Case(C) + Case(D)

=7 + 84 + 210 + 140

=441

(iii) At least three girls

Total number of ways the team can have at least three girls =⁴C_3. ⁷C₂+⁴C_4.⁷C₁

= 4 × 21 + 7

= 84 + 7

= 91

Objective Type Questions

Choose the correct answer out of the given four options against each of the Exercises from 26 to 40 (M.C.Q.).

A. 20

Explanation:

According to the question,

ⁿC₁₂ = ⁿC₈

We know that,

ⁿC_r

n – 8 = 12

n = 20

Hence, Option (A) 20 is the correct answer.

B. 64

Explanation:

The coin is tossed 6 times.

Number of possible outcomes = 2

The number of possible outcomes when a coin is tossed 6 times is 2⁶ = 64

Hence, Option (B) 64 is the correct answer.

C. 24

Explanation:

We know that,

ⁿP_r=

Four-digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once

=⁴P₄ = 4! = 24

Hence, Option (C) 24 is the correct answer.

B. 108

Explanation:

The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time =(3+4+5+6)3! =108

Hence, Option (B) 108 is the correct answer.

C. 7200

Explanation:

We know that,

ⁿC_r

Total number of given vowels=4

Total number of given consonant=5

Total number of words that can be formed by 2 vowels and 3 consonants =⁴C₂.⁵C_{3 =} 4!/2!2!

2 vowels and 3 consonants =5!

Total number of word = 5! ×4!/2!2! = 7200

Hence, Option (C) 7200 is the correct answer.