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June 2019 p63 q4
2824
(i) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if there must be at least twice as many men as there are women on the committee.
(ii) Find the number of ways a committee of 6 people can be chosen from 8 men and 4 women if 2 particular men refuse to be on the committee together.
Solution
(i) To satisfy the condition of having at least twice as many men as women, consider the following cases:
4 men and 2 women: The number of ways is \(\binom{8}{4} \times \binom{4}{2} = 70 \times 6 = 420\).
5 men and 1 woman: The number of ways is \(\binom{8}{5} \times \binom{4}{1} = 56 \times 4 = 224\).
6 men and 0 women: The number of ways is \(\binom{8}{6} \times \binom{4}{0} = 28 \times 1 = 28\).
\(Total number of ways = 420 + 224 + 28 = 672.\)
(ii) First, calculate the total number of ways to choose 6 people from 12 (8 men + 4 women):
\(\binom{12}{6} = 924\).
Now, calculate the number of ways where the two particular men are together. Treat them as a single unit, so choose 4 more people from the remaining 10:
\(\binom{10}{4} = 210\).
Subtract the cases where the two men are together from the total:
