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?

  • 1
    $\begingroup$ Mqtt's answer is write, but it should also be noted that you want ${}_{17}C_3$, since there are only $17$ women to choose from, not ${}_{38}C_3$ $\endgroup$ – Thomas Andrews Nov 16 '14 at 20:34

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

  • $\begingroup$ For the 3rd and 4th problem I got values that are larger than the total number of committees possible using (for example #4) $_{17}C_1 \bullet _{37}C_6$. Do you have any hints as to what I'm doing wrong here? $\endgroup$ – hax0r_n_code Nov 17 '14 at 1:15
  • $\begingroup$ Please see my edit above. $\endgroup$ – Matt Samuel Nov 17 '14 at 1:20
  • $\begingroup$ Ahhhhhhhh! Ok, so number 4 makes perfect sense now. Thank you! I was able to figure out the first question (but thank you as well for the help there). I'll try the 3rd and show my results here. $\endgroup$ – hax0r_n_code Nov 17 '14 at 1:25
  • $\begingroup$ For the 3rd I got $_{17}C_{3} \bullet _{35}C_{4}$ $\endgroup$ – hax0r_n_code Nov 17 '14 at 1:41
  • $\begingroup$ The third one is a bit more complicated, see my answer above. You need to split it up into cases. $\endgroup$ – Matt Samuel Nov 17 '14 at 1:50

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"

  • $\begingroup$ Thank you! I understand the first one, but the second I assume would follow the addition principle? Does the second one seems correct, but is it? $\endgroup$ – hax0r_n_code Nov 16 '14 at 20:42
  • $\begingroup$ The second one is indeed correct. It uses the addition principle since there is absolutely no chance for overlap and we want to count either situation. The third and fourth will need the multiplication principle. $\endgroup$ – JMoravitz Nov 16 '14 at 20:45
  • $\begingroup$ For the 3rd one I have $_{17}C_{3} \bullet _{35}C_{4}$, but this number is larger than the total number of committees possible (regardless of men or women chosen). So this seems like it is incorrect to me? $\endgroup$ – hax0r_n_code Nov 16 '14 at 20:51
  • $\begingroup$ For the 4th one I have $_{17}C_1 \bullet _{37}C_6$ and this too is larger than the total number of committees possible. Any hint as to what I'm doing wrong here? $\endgroup$ – hax0r_n_code Nov 17 '14 at 1:12
  • $\begingroup$ for the fourth pick the one woman (by doing $\binom{17}{1}$) and then pick which of the men they are, (by doing $\binom{21}{6}$). You wound up doing 37choose6 instead of 21choose6. Remember that the six remaining people must necessarily be men since we are restricted to having only one woman (so none of the other women can be selected for the remaining six spots) $\endgroup$ – JMoravitz Nov 17 '14 at 2:00

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