Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4

Question 1.

If ^{n}C_{8} = ^{n}C_{2}. Find ^{n}C_{2}.

Solution:

We have: ^{n}C_{r} = ^{n}C_{n-r}

^{n}C_{2} = ^{n}C_{n-2}

^{n}C_{8} = ^{n}C_{n-2} ⇒ n – 2 = 8 or n = 10.

^{n}C_{2} = ^{10}C_{2} = \(\frac{10×9}{1×2}\) = 45.

Question 2.

Determine n, if

(i) ^{2n}C_{3} : ^{n}C_{2} = 12 : 1

(ii) ^{2n}C_{3} : ^{n}C_{3} = 11 : 1

Solution:

^{2n}C_{3} : ^{n}C_{2} = 12 : 1

2n – 1 = 9, 2n = 10 or n = 5.

(ii) ^{2n}C_{3} : ^{n}C_{3} = 11 : 1

∴ 4(2n – 1) = 11(n – 2)

or 8n – 4 = 11n – 22

or 3n = 18 ⇒ n = 6.

Question 3.

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

Solution:

We get a chord by joining two points. If p is the number of chords from 21 points, then

p = C(21, 2) = \(\frac{21!}{2!(21 – 2)!}\) = \(\frac{21!}{2!19!}\) = \(\frac{21×20(19)!}{2×1(19!)}\)

= 21 × 10 = 210 chords.

Question 4.

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

Solution:

3 boys out of 5 boys can be selected in ^{5}C_{3} ways.

3 girls out of 4 girls can be selected in ^{4}C_{3} ways.

∴ Number of ways in which 3 boys and 3 girls can be selected

= ^{5}C_{3} × ^{4}C_{3} = ^{5}C_{2} × ^{4}C_{1}.

= \(\frac{5×4}{1×2}\) × \(\frac{4}{1}\) = 40. [∵ ^{n}C_{2} = ^{n}C_{n-2}.

Question 5.

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?

Solution:

The number of ways of selecting 3 red balls out of 6 red balls = ^{6}C_{3}.

The number of ways of selecting 3 white balls out of 6 white balls = ^{5}C_{3}.

The number of ways of selecting 3 blue balls out of 5 blue balls = ^{5}C_{3}.

The number of ways of selecting 3 balls of each colour

= ^{6}C_{3} × ^{5}C_{3} × ^{5}C_{3} = ^{5}C_{3} × ^{5}C_{2} × ^{5}C_{2}.

= \(\frac{6×5×4}{1×2×3}\) × \(\frac{5.4}{1×2}\) × \(\frac{5.4}{1.2}\).

= 20 × 10 × 10 = 2000.

Question 6.

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

Solution:

One ace will be selected from four aces and four cards will be selected from (52 – 4 =) 48 cards. If p is the required number of ways, then

p = C(4, 1) × C(48, 4)

= 4 × 2 × 47 × 46 × 45 = 778320 ways.

Question 7.

In how many ways, can we select a cricket eleven from 17 players in which only 5 players can bowl, if each cricket eleven must include exactly 4 bowlers?

Solution:

Four bowlers can be selected from the five bowlers and seven players can be selected from 12 players (17 – 5 = 12). If ‘p’ is the number of ways of selecting the cricket eleven, then

p = C(5, 4) × C(12, 7)

= 3960.

Question 8.

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

Solution:

The number of ways in which 2 black balls out of 5 black balls can be selected = ^{5}C_{2}.

The number of ways in which 3 red balls out of 6 red balls can be selected = ^{6}C_{3}.

The number of ways in which 2 black and 3 red balls can be selected

= ^{5}C_{2} × ^{6}C_{3} = \(\frac{5×4}{1×2}\) × \(\frac{6×5×4}{1×2×3}\).

= 10 × 20 = 200.

Question 9.

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

Solution:

Out of available nine courses, two are compulsory. Hence, the student is free to select 3 courses out of 7 remaining courses. If p is the number of ways of selecting 3 courses out of 7 courses, then

p = C(7, 3)

= \(\frac{7!}{3!(7 – 3)!}\) = \(\frac{7!}{3!4!}\) = \(\frac{7×6×5×4!}{3×2×1×4!}\)

= 7 × 5 = 35 ways.