Question-1 :- If nC8 = nC2, find nC2.
Solution :-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?
Solution :-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?
Solution :-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.
Solution :-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.
Solution :-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?
Solution :-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.
Solution :-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?
Solution :-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.
