<s>[INST] From 7 men & 4 women, 4 are to be selected to form a committee so that at least a woman is there on the committee. In how many ways can it be done? 
From $7$ men & $4$ women, $4$ people are to be selected to form a committee so that at least a woman is there on the committee. In how many ways can it be done?

I was trying this in the following way, but surely I am missing something:
As there must be $1$ woman, there are $\binom{4}{1}$ (using the other notation this is $C_4^1$) ways to select $1$ woman from $4$
There are $\binom{10}{3}$ ( or$C_{10}^3$) ways to select $3$ people from the remaining $10$.
So $\binom{4}{1}\times \binom{10}{3} = 480$ ways.
 [/INST] There are 7 men and 4 women, and we need 4 members on the committee such that at least there is one woman on it. In that regard, we can make the following combinations.
We can have 4 women and 0 men. Or we can have 3 women and 1 man. Or we have 2 women and 2 men. Or we can have 1 woman and 3 men. We have to stop at this point because we need at least one woman on the committee. Start with the first case, where we have 4 women and 0 men. There $\binom{4}{4}\times\binom{7}{0}$ ways to form the committee.
In the second case where we have 3 women and 1 man, there are $\binom{4}{3}\times\binom{7}{1}$ ways to form the committee. In the third case, there are $\binom{4}{2}\times\binom{7}{2}$ ways to form the committee. In the last case, there are $\binom{4}{1}\times\binom{7}{3}$ ways to form the committee. To find the total number of ways to form the 4-member committee, we need to add the ways from the individual cases. Therefore, there are
$$\binom{4}{4}\times\binom{7}{0}+\binom{4}{3}\times\binom{7}{1}+\binom{4}{2}\times\binom{7}{2}+\binom{4}{1}\times\binom{7}{3}\\ =1+28+126+140\\=295$$ ways to form the committtee.
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