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Feb/Mar 2016 p62 q6
2842
Hannah chooses 5 singers from 15 applicants to appear in a concert. Of the 15 applicants, 10 are female and 5 are male. Hannah's friend Ami would like the group of 5 performers to include more males than females.
(iii) Find the number of different selections of 5 performers with more males than females.
(iv) Two of the applicants are Mr and Mrs Blake. Find the number of different selections that include Mr and Mrs Blake and also fulfil Ami’s requirement.
Solution
(iii) To have more males than females in a group of 5, we can have the following combinations:
3 males and 2 females: The number of ways to choose 3 males from 5 is \(\binom{5}{3}\), and the number of ways to choose 2 females from 10 is \(\binom{10}{2}\). Thus, the total number of ways is \(\binom{5}{3} \times \binom{10}{2} = 450\).
4 males and 1 female: The number of ways to choose 4 males from 5 is \(\binom{5}{4}\), and the number of ways to choose 1 female from 10 is \(\binom{10}{1}\). Thus, the total number of ways is \(\binom{5}{4} \times \binom{10}{1} = 50\).
5 males and 0 females: The number of ways to choose 5 males from 5 is \(\binom{5}{5}\), and the number of ways to choose 0 females from 10 is \(\binom{10}{0}\). Thus, the total number of ways is \(\binom{5}{5} \times \binom{10}{0} = 1\).
Adding these, the total number of ways is \(450 + 50 + 1 = 501\).
(iv) Mr and Mrs Blake are already chosen, so we need to choose 3 more performers. We need more males than females, so the combinations are:
3 males and 0 females: Choose 3 males from the remaining 4 males. The number of ways is \(\binom{4}{3} \times \binom{9}{0} = 4\).
2 males and 1 female: Choose 2 males from the remaining 4 males and 1 female from 9 females. The number of ways is \(\binom{4}{2} \times \binom{9}{1} = 54\).
Adding these, the total number of ways is \(4 + 54 = 58\).
