How many committees are possible if a committee must have $3$ women and $4$ men?
$_{38}C_3+_{38}C_4$ or $\frac{38!}{3!35!}+\frac{38!}{4!34!} = 8,435+73,815 = 82,251$
How many committees are possible if a committee must consist of all women or all men?
$_{17}C_7+_{21}C_7$ or $\frac{17!}{7!10!}+\frac{21!}{7!14!} = 19,448+116,280 = 135,728$
How many committees are possible if a committee must have at least 3 women?
$_{38}C_3+_{35}C_4$ or $\frac{38!}{3!35!}+\frac{35!}{4!31!} = 8,435+52,360 = 60,795$
How many committees are possible if a committee must have exactly one woman?
$_{38}C_1+_{37}C_6$ or $\frac{38!}{1!37!}+\frac{37!}{6!31!} = 38+2,324,821 = 2,324,821$
Am I going about these the correct way?
Best Answer
If we want to compute how many sets there are consisting of $m$ men and $n$ women, the correct answer is given by multiplying the number of combinations.
For example, for the first question we want 3 women and 4 men. There are 17 women and we want to choose 3 of them, so that is $\binom{17}{3}$. The number of sets of 4 men is $\binom{21}{4}$. Therefore the answer is $\binom{17}{3}\binom{21}{4}$.
For the third question we will need to do some adding. If we choose a set of three women and then arbitrarily choose a set of 4 from the remaining people, we will count some choices more than once. Therefore we will consider the cases where there are 3, 4, 5, 6, and 7 women separately so we can control the sets we are picking. For that we get $$\binom{17}{3}\binom{21}{4}+\binom{17}{4}\binom{21}{3}+\binom{17}{5}\binom{21}{2}+\binom{17}{6}21+\binom{17}{7}\mathrm{.}$$
For the fourth question, we first want to enumerate the number of sets with exactly 1 woman. Since there are 17 women, there are exactly 17 sets consisting of just one woman. The remaining people must be men, so we need to choose 6 out of 21 men, and there are $\binom{21}{6}$ ways to do this. The answer is therefore $17\binom{21}{6}=\binom{17}{1}\binom{21}{6}$.