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Say I have a set $\{A, B, C, D, E, F\}$ and I have to find how many sets of four elements I can make from these that must include at least any two elements from the set $\{D, E, F\}$?

On a similar basis: How many ways can you form a committee of five from $6$ men and $4$ women such that in every of these committees there are at least $2$ women?

I seriously have no idea how to go about these.

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up vote 2 down vote accepted

In both of these problems it’s easiest to count the sets that don’t meet the requirement and subtract them from the total number of possible sets.

In the second problem, for instance, there are $\binom{10}5$ possible $5$-person committees altogether. $\binom65$ of them include no women, and $\binom64\binom41$ include just one woman, so there are $$\binom{10}5-\binom65-\binom64\binom41$$ committees that include at least two women. The first problem can be worked in similar fashion.

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Thank you for the answer, but I think I should have made my answer more general: What if I had 10000 men and 1000 women, and I have to form a committee of 100 in which at least 20 should be women? It seems like an impossible task to go about the individual cases here. – Alraxite Feb 6 '13 at 8:58
@Alraxite: I’m fairly sure that getting an exact result would simply be computationally ugly. – Brian M. Scott Feb 6 '13 at 9:04
Okay, and would the answer to the first problem be $\binom64- \binom44 \binom31$? – Alraxite Feb 6 '13 at 9:30
@Alraxite: Yes, though I’d get it simply by noticing that there are exactly $3$ possible $4$-person committees that don’t meet the requirement and writing $\binom64-3$ directly. – Brian M. Scott Feb 6 '13 at 9:31
@Alraxite: Oops: I didn’t read carefully enough. You want $\binom64-\binom33\binom31$: the $\binom33$ is for taking all $3$ of A, B, and C. (Your $\binom44$ was probably just a typo, but I thought that I’d better make sure.) – Brian M. Scott Feb 6 '13 at 9:36

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