TOPICS

Exercise - 7.4

Permutations and Combinations

**Question-1 :-** If ⁿC₈ = ⁿC₂, find ⁿC₂.

ⁿC₈ = ⁿC₂ We know that, ⁿCᵤ = ⁿCᵥ; u = v or n = u + v So, n = 8 + 2 = 10 Therefore, ¹⁰C₂ = 10!/2!8! = 45

**Question-2 :-** Determine n if

(i) ²ⁿC₃ : ⁿC₃ = 12 : 1

(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1

(i) ²ⁿC₃ : ⁿC₃ = 12 : 1 (2n)!/[3!(2n-3)!] : n!/[3!(n-3)!] = 12 : 1 [2n x (2n-1) x (2n-2)]/3! : [n x (n-1) x (n-2)]/3! = 2 : 1 [4(2n-1)(n-1)]/[(n-1)(n-2)] = 12 (2n-1)/(n-2) = 3 2n - 1 = 3n - 6 3n - 2n = -1 + 6 n = 5

(ii) ²ⁿC₃ : ⁿC₃ = 11 : 1 (2n)!/[3!(2n-3)!] : n!/[3!(n-3)!] = 11 : 1 [2n x (2n-1) x (2n-2)]/3! : [n x (n-1) x (n-2)]/3! = 11 : 1 [4(2n-1)(n-1)]/[(n-1)(n-2)] = 11 (8n-4)/(n-2) = 11 8n - 4 = 11n - 22 11n - 8n = -4 + 22 3n = 18 n = 6

**Question-3 :-** How many chords can be drawn through 21 points on a circle?

For drawing one chord on a circle, only 2 points are required. To know the number of chords that can be drawn through the given 21 points on a circle, the number of combinations have to be counted. Therefore, there will be as many chords as there are combinations of 21 points taken 2 at a time. Thus, required number of chords = ²¹C₂ = 21!/2!19! = 210

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

A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls. 3 boys can be selected from 5 boys in ⁵C₃ ways. 3 girls can be selected from 4 girls in ⁴C₃ ways. Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected = ⁵C₃ x ⁴C₃ = 10 x 4 = 40

**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.

There are a total of 6 red balls, 5 white balls, and 5 blue balls. 9 balls have to be selected in such a way that each selection consists of 3 balls of each colour. 3 balls can be selected from 6 red balls in ⁶C₃ ways. 3 balls can be selected from 5 white balls in ⁵C₃ ways. 3 balls can be selected from 5 blue balls in ⁵C₃ ways. Thus, by multiplication principle, required number of ways of selecting 9 balls = ⁶C₃ x ⁵C₃ x ⁵C₃ = 20 x 10 x 10 = 2000

**Question-6 :-** Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

In a deck of 52 cards, there are 4 aces. A combination of 5 cards have to be made in which there is exactly one ace. Then, one ace can be selected in ⁴C₁ ways and the remaining 4 cards can be selected out of the 48 cards in ⁴⁸C₄ ways. Thus, by multiplication principle, required number of 5 card combinations = ⁴⁸C₄ x ⁴C₁ = 778320

**Question-7 :-** 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?

Out of 17 players, 5 players are bowlers. A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers. 4 bowlers can be selected in ⁵C₄ ways and the remaining 7 players can be selected out of the 12 players in ¹²C₇ ways. Thus, by multiplication principle, required number of ways of selecting cricket team = ⁵C₄ x ¹²C₇ = 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 can be selected.

There are 5 black and 6 red balls in the bag. 2 black balls can be selected out of 5 black balls in ⁵C₂ ways and 3 red balls can be selected out of 6 red balls in ⁶C₃ ways. Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls = ⁵C₂ x ⁶C₃ = 10 x 20 = 200.

**Question-9 :-** 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?

There are 9 courses available out of which, 2 specific courses are compulsory for every student. Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be chosen in ⁷C₃ ways. Thus, required number of ways of choosing the programme = ⁷C₃ = 7!/3!4! = 35.

CLASSES