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What's the number of ways to choose a 6-member team of at least 4 women, if we choose from the pool of 8 women and 6 men?

I was thinking it's $\mathrm{C}_8^4\cdot\mathrm{C}_{10}^2$ but it's actually way more than the maximum possible number of $\mathrm{C}_{14}^6$.

How should I reason this?

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  • $\begingroup$ There are some double (or more) counting. The case where more than 4 women are in the team are counted more than once. Says, A, B, C and D are first chosen via the first term and then E is chosen in the second term. Suppose the men are unchanged. When the first four are any combination of these five and the next one is also in these five then the combination is counted more than once. That's why the number is more than the maximum. $\endgroup$ – Karn Watcharasupat Nov 23 '17 at 18:20
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Consider choosing $n$ women. Clearly the number of men is $6-n$.
So we have the total number of ways as $$\sum_{n=4}^6 {8 \choose n}{6 \choose {6-n}}=1414$$ so that we account for all cases in one go.

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