Combinations - 17 women and 21 men to form a committee of size 7

Question

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?

Answer 1

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 $17\binom{21}{6}=\binom{17}{1}\binom{21}{6}$.

Answer 2

It seems that you are forgetting the limitations on how many women and men there are. Also as Matt mentioned above me, this relies on the rule of product (a.k.a. the multiplication principle) so you will multiply in most of these scenarios, not add.

For the first problem, you ask the question, "Which 3 of the 17 women are in my committee" followed by "Which 4 of my 21 men are in my committee" for an answer of $\binom{17}{3}\cdot\binom{21}{4}$. Which is different than $\binom{38}{3}\cdot\binom{38}{4}$

In the 38choose3 interpretation it is wrong for two reasons, you might have chosen the same person in both the first and second counting, and furthermore, the distinction would have nothing to do with their gender. $\binom{38}{3}\cdot\binom{38}{4}$ would however count a situation such as "With 38 people in the red team and 38 people in the blue team, how many ways can we have three red players and 4 blue players be selected to participate in a group"