A committee of 6 people is to be chosen from 9 women and 5 men.
(a) Find the number of ways in which the 6 people can be chosen if there must be more women than men on the committee.
The 9 women and 5 men include a sister and brother.
(b) Find the number of ways in which the committee can be chosen if the sister and brother cannot both be on the committee.
Solution
(a) To have more women than men, consider the following scenarios:
- 6 women and 0 men: \(\binom{9}{6} = 84\)
- 5 women and 1 man: \(\binom{9}{5} \times \binom{5}{1} = 126 \times 5 = 630\)
- 4 women and 2 men: \(\binom{9}{4} \times \binom{5}{2} = 126 \times 10 = 1260\)
\(Total number of ways = 84 + 630 + 1260 = 1974.\)
(b) Total number of ways to choose 6 people from 14: \(\binom{14}{6} = 3003\).
Number of ways with both sister and brother: \(\binom{12}{4} = 495\).
Number of ways without both sister and brother: \(3003 - 495 = 2508\).
Alternatively, calculate:
- Number of ways with neither: \(\binom{12}{6} = 924\)
- Number of ways with either brother or sister (not both): \(\binom{12}{5} \times 2 = 792 \times 2 = 1584\)
\(Total number of ways = 924 + 1584 = 2508.\)
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