<s>[INST] Choosing at least 2 women from 7 men and 4 women In how many different ways can we choose six people, including at least two women, from a group made up of seven men and four women?
Attempt:
As we have to have at least two women in the choices, then $\displaystyle\binom{4}{2}$, leaving a total of $4$ out of $9$ people to be chosen. So $$\binom{9}{4} \cdot \binom{4}{2}=756$$
The answer is $371$. Where is my error?
 [/INST] Consider three cases:

*

*choosing 2 women and 4 men

*choosing 3 women and 3 men

*choosing 4 women and 2 men

So the total number of arrangements is:
$\binom{4}{2} \times \binom{7}{4}+ \binom{4}{3} \times \binom{7}{3}+ \binom{4}{4} \times \binom{7}{2}=371$
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