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Suppose that a department contains 10 men and 15 women. How many ways are there to form a committee with six members if it must have at least as many women as there are men in it?

So my current approach on this question is to make 4 different cases where there are at least as many women as there are men. So I have:

Case 1: 6 Women, 0 Men

Case 2: 5 Women, 1 Men

Case 3: 4 Women, 2 Men

Case 4: 3 Women, 3 Men

For each case I have calculated the combinations as follows:

Case 1: $\binom{15}{6}$

Case 2: $\binom{15}{5} * \binom{10}{1}$

Case 3: $\binom{15}{4} * \binom{10}{2}$

Case 4: $\binom{15}{3} * \binom{10}{3}$

I'm not sure what I should do at this point or even if what I have done is correct.

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    $\begingroup$ Everything you've done looks fine, just add up the cases and you're done. $\endgroup$ – user2566092 Mar 9 '15 at 23:50
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This is a great start! In each case you have counted the number of ways you could form a group with a fixed number of women and men (with the former at least as large as the latter). Now sum up your different cases in order to find the total!

You may want to evaluate your 'choose' functions (binomial coefficients) using the following formula: $$ {n \choose k} = { n! \over k! (n-k)! } $$

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