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# NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination

NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination are available here for boosting the preparation of class 11 maths exam. The NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutaion and Combination are created by an expert of maths who has huge experience in teaching maths from class 9 to class 12.All the questions of permutation and combination are solved by step by step ways therefore every students of the grade 11 can understand the solutions easily .

## NCERT Solutions for Class 11 Maths  Chapter 7 Permutation and Combination

Exercise 7.1- Permutations and Combinations

Exercise 7.2 -Permutation and Combination

Exercise 7.3-Permutation and Combination

Exercise 7.4-Permutation and Combination

Miscellaneous Exercise-Permutation and Combination

## NCERT Solutions for Class 11 Maths Exercise 7.4 of the Chapter 7 Permutation and Combination

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Q1. If nC8 = nc2, find  nc2

Ans. Since nCr =nCn-r

nC8 = nCn-8,

Substituting nC8 = nCn-8,

nCn-8 = nc2

n – 8 = 2⇒ n = 10

$\fn_cm ^{n}c_{r}=\frac{n!}{r!\left ( n-r \right )!}$

$\fn_cm ^{10}c_{2}=\frac{10!}{2!\left ( 10-2 \right )!}=\frac{10\times 9\times 8!}{2\times 8!}=45$

Hence the required value of  nc2is 45

Q2.Determine n if

(i) 2nC3: nc3 = 12 : 1

(ii) 2nC3: nc3 = 11 : 1

Ans. It is given to us that

2nC3: nc3 = 12 : 1

$\fn_cm \Rightarrow \frac{\left ( 2n \right )!}{3!\left ( 2n-3 \right )}:\frac{n!}{3!\left ( n-3 \right )!}=12:1$

$\fn_cm \Rightarrow \frac{2n\left ( 2n-1 \right )\left ( 2n-2 \right )\left ( 2n-3 \right )!}{3!\left ( 2n-3 \right )}:\frac{n\left ( n-1 \right )\left ( n-2 \right )\left ( n-3 \right )!}{3!\left ( n-3 \right )!}=12:1$

4(2n -1)(n -1) : (n -1)(n -2) = 12 : 1

$\fn_cm \frac{4\left ( 2n-1 \right )}{n-2}=\frac{12}{1}$

8n -4 = 12n – 24

4n = 20

n = 5

(ii) 2nC3: nc3 = 11 : 1

Similarly as in (i),we have

$\fn_cm \frac{4\left ( 2n-1 \right )}{n-2}=\frac{11}{1}$

11n – 22 = 8n – 4

3n = 18

n = 6

Q3. How many chords can be drawn through 21 points on a circle?

Ans.For drawing chords we are required to choose two points out of 21 points on the circle, therefore the number of ways of such combinations are  21C2

The number of chords are = 21C2

$\fn_cm =\frac{21!}{2!\left ( 21-2 \right )!}=\frac{21\times 20\times 19!}{2\times 19!}=210$

Hence total number of chords can be drawn are  210

Q4.In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls.

Ans. The number of ways 3 boys can be selected out of 5 boys are 5C3

The number of ways 3 girls can be selected out of 4 girls are 4C3

According to the multiplication rule,total number of ways selecting  a team of 3 boys and 3 girls are = 5C3 ×4C3

$\fn_cm =\frac{5!}{3!\left ( 5-3 \right )!}\times \frac{4!}{3!\left ( 4-3 \right )!}$

$\fn_cm =\frac{5!}{3!\left ( 5-3 \right )!}\times \frac{4!}{3!\left ( 4-3 \right )!}=\frac{5\times 4\times 3!\times 4\times 3!}{3!\times 3!\times 2}=40$

Hence the number of ways  a team of 3 boys and 3 girls  selected from 5 boys and 4 girls is 40

Q5.Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Ans. The number of ways of selecting 3 balls out of 6 red balls are  6C3

The number of ways of selecting 3 balls out of 5 white balls are  5C3

The number of ways of selecting 3 balls out of 5 blue balls are  5C3

According to the multiplication rule,total number of ways selecting  9 balls out of 6 red ball,5 white balls and 5 blue balls are = 6C3 ×5C×5C3

$\fn_cm =\frac{6!}{3!\left ( 6-3 \right )!}\times \frac{5!}{3!\left ( 5-3 \right )!}\times \frac{5!}{3!\left ( 5-3 \right )!}$

$\fn_cm =\frac{6\times 5\times 4\times 3!}{3\times 2\times 3!}\times \frac{5\times 4\times 3!}{3!\times 2}\times \frac{5\times 4\times 3!}{3!\times 2}$

= 20 × 10 × 10 = 2000

Hence required number of ways of selecting 9 balls are 2000

Q6.Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Ans. The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination are =The number of ways selecting 1 ace out of 5 cards × The number of ways selecting remaining 4 cards out of remaining (52 -4= 48 ) cards

The number of ways selecting 1 ace out of 4 cards = 4C1

The number of ways selecting  remaining 4 cards out of  remaining (52 -4 = 48) cards = 48C4

The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination are = 4C1 × 48C4

$=\frac{4!}{1!3!}\times \frac{48!}{4!\times 44!}$

$=\frac{4\times 3!}{1!\times 3!}\times \frac{48\times 47\times 46\times 45\times 44!}{4\times 3\times 2\times 44!}$

= 4× 2× 47 ×46 ×45 = 778320

The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination is 778320

Q7.In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers.

Ans.The number of ways selecting  a team of 11 players = The number of ways selecting 4 bowlers out of 5 bowlers × The number of ways selecting 7 players out of remaining (17-5 =12) players

The number of ways selecting  a team of 11 players = 5C4×12C7

$\fn_cm =\frac{5!}{4!\times 1!}\times \frac{12!}{7!\times 5!}$

$\fn_cm =\frac{5\times 4!}{4!}\times \frac{12\times 11\times 10\times 9\times 8\times 7!}{7!\times 5\times 4\times 3\times 2}$

=5 × 11×9×8 = 3960

The number of ways selecting  a team of 11 players in which there are exactly 4 bowlers is  3960

Q8.A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Ans.The number of ways selecting 2 black balls out of 5 black balls = 5C2

The number of ways selecting 3 red balls out of 6 red balls = 6C3

The number of ways selecting 2 black balls and 3 red balls out of 5 black balls and 6 red balls =5C×  6C3

$\fn_cm =\frac{5!}{2!\times 3!}\times \frac{6!}{3!\times 3!}$

$\fn_cm =\frac{5\times 4\times 3!}{2\times 3!}\times \frac{6\times 5\times 4\times 3!}{3\times 2\times 3!}$

= 10 × 20 = 200

The number of ways selecting 2 black balls and 3 red balls out of 5 black balls and 6 red balls is 200

Q9.In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Ans. It is given that student has to choose 5 coarses out of 9 coarses in which 2 coarses are compulsury.

Since 2 coarses are compusury therefore number of ways selecting 5 coarses = Number of ways of selecting  remaining (5 -2 = 3)  coarses out of remaining (9 -2 = 7) coarses is = 7C3

$\fn_cm \frac{7!}{3!\times \left ( 7-3 \right )!}=\frac{7\times 6\times 5\times 4!}{3\times 2\times 4!}=35$

Hence required number of ways selecting 5 coarses is 35

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