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.
 A: This is not good. Let the women be $w_1,..,w_4$ and men $m_1,...,m_7$. You choose $1$ woman, say $w_1$. Then you chose $3$ people form the $10$ remaining, say $w_3,m_2,m_7$. But this is the equivalent to choosing $w_3$ and then, while choosing from the reemaining $10$ getting $w_1,m_2,m_7$ So this doesn't work. What i suggest you do is subtract the number of gropus of $4$ with no women in it (which is $\binom{7}{4}$, we only vhoose from the $7$ men) from the total number of possible gropus (which is $\binom{11}{4}$).
So we have $$\binom{11}{4}-\binom{7}{4}=330-35=295$$
ways of choosing people.
P.S. try to avoid $C_a^b$. try to use $\binom{a}{b}$ instead.
A: Let me ask you this. If we had to select $2$ people from a group of $3$ would you first select one in $\binom{3}{1}$ and the next in $\binom{2}{1}$ ways to total $\binom{3}{1}\binom{2}{1}$ ways or both in $\binom{3}{2}$ ways? The former will give us $P_{2}^3$ instead. Can you now see the mistake?
A: 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.
A: Actually there is a more elegant answer.
We just count the number of ways of choosing a committee without a man-woman restriction and then subtract the ways in which only men are selected. These are called complementary events. The answer, thus is ${11 \choose 4}-{7 \choose 4}=295$. ($11$ is the total number of people and $7$ is the number of men).
