AnswersRelated In how many ways a committee consisting of 5 men and 3 women, can be chosen from 9 men and 12 women.a)12459b)75200c)27720d)18920Correct answer is option 'C'. Can you explain this answer?
Choose 5 men out of 9 men = 9C5 ways = 126 ways Choose 3 women out of 12 women = 12C3 ways = 220 ways The committee can be chosen in 27720 ways
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If we want to compute how many sets there are consisting of $m$ men and $n$ women, the correct answer is given by multiplying the number of combinations. For example, for the first question we want 3 women and 4 men. There are 17 women and we want to choose 3 of them, so that is $\binom{17}{3}$. The number of sets of 4 men is $\binom{21}{4}$. Therefore the answer is $\binom{17}{3}\binom{21}{4}$. For the third question we will need to do some adding. If we choose a set of three women and then arbitrarily choose a set of 4 from the remaining people, we will count some choices more than once. Therefore we will consider the cases where there are 3, 4, 5, 6, and 7 women separately so we can control the sets we are picking. For that we get $$\binom{17}{3}\binom{21}{4}+\binom{17}{4}\binom{21}{3}+\binom{17}{5}\binom{21}{2}+\binom{17}{6}21+\binom{17}{7}\mathrm{.}$$ For the fourth question, we first want to enumerate the number of sets with exactly 1 woman. Since there are 17 women, there are exactly 17 sets consisting of just one woman. The remaining people must be men, so we need to choose 6 out of 21 men, and there are $\binom{21}{6}$ ways to do this. The answer is therefore $17\binom{21}{6}=\binom{17}{1}\binom{21}{6}$.
The only problem with multiplying by seven is that you could have one case where you fix one woman and choose other members in such a way so that the people on the committee are the same as in another case where you fix another woman. To make this clearer, let W1, W2, ..., W7 be the women who can be chosen to be on the committee, and let M1, M2, ..., M9 be the possible men. For example, you would be counting the case where you fix W1 and choose W2, W3, M1, and M2 separately from the case where you fix W2, and choose W1, W3, M1, and M2, even though they form the same committees. I also don't think the correct answer is C(15,4), because the choice of which woman to fix is important as well. For example, counting the number of possibilities as C(15,4) would count all the cases where you choose one woman and the rest men as one case, even though these are in fact seven separate cases. I would rather simply count the total number of committees and then subtract the number that don't have any women (i.e. with all men) to get the number with at least one woman on the committee. This should come out with C(16,5)-C(9,5) as your answer. |
How many ways can a committee be formed of 5 men and 3 women picked from a pool of 9 men and 12 women?
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