# How to find the number of combinations in which a class of elements always has to be included?

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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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.
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