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.
Best Answer
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.