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For this question:

A committee of six is to be selected from a group of ten men and 12 women. In how many ways can the committee can be chosen if it has to contain at least two men and one woman?

Working I have attempted so far:

1.)$$\binom{22}{6}-\left(\binom{10}{6}+\binom{10}{5}\binom{12}{1}+\binom{12}{6}\right)=70959 $$

So basically I am not really sure if it is right, It seems right to me though. Please help in correcting my working if wrong

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    $\begingroup$ Your work is correct since you subtracted the number of committees that were not allowed from the total possible ways of selecting the committees. $\endgroup$ – N. F. Taussig Apr 10 '15 at 12:34
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    $\begingroup$ It should be $\binom {12}5\binom{10}1$ instead of $\binom {10}5\binom{12}1$. $\endgroup$ – Hypergeometricx Apr 10 '15 at 16:54
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you want to have at least 2 men and 1 woman in the committee, so you have to subtract the ways that all of the committee are men, all of the committee are women and also the committee has just 1 man and 5 women. the answer is: $\\$

$$\binom{22}{6}-\left(\binom{10}{6}+\binom{10}{1}\binom{12}{5}+\binom{12}{6}\right)$$

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  • $\begingroup$ ahhh thanks for explanation guys, did not realize my mistake $\endgroup$ – Itakura Apr 11 '15 at 11:14
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Let (m, w), where m = no. of men forming the committee and w = no. of women forming the committee, denote the number of committee formed.

Choosing at least 2 men and 1 woman = C(22, 6) - (0,6) - (1,5) - (6,0)

Therefore the number of ways = C(22, 6) - C(12, 6) - C(10, 1)*C(12, 5) - C(10, 6) = 65559.

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Since the committee must have between 2 and 5 men, an alternate method

(which involves more calculation) is to take

$\displaystyle\binom{10}{2}\binom{12}{4}+\binom{10}{3}\binom{12}{3}+\binom{10}{4}\binom{12}{2}+\binom{10}{5}\binom{12}{1}=65,559$.

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