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Say I have a class of $24$ students. $16$ are men and $8$ are women. And I want to choose a team of $8$ that has four men and three women, the eighth member can be anyone. Then I would think that there are $$\binom{16}{4}\cdot \binom{8}{3}\cdot \binom{17}{1} = 1732640$$ different possible teams that meet this requirement. That is first choosing the four men, then the three women, and then the one random.

But then I noticed that there are only $$\binom{24}{8} = 735471$$ different groups of eight in general. So how am I pulling extra mythical groups? Any help would be appreciated.

Dan
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Chris Christopherson
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  • Essentially, you need to count the pairs $(V,W)$ of subsets such that $V\subseteq M$ and $W\subseteq X\setminus M$ and $\lvert V\rvert+\lvert W\rvert=8$ and [$\lvert V\rvert=4$ or $\lvert V\rvert=5$]. You are counting the triples $(A,B,C)$ of subsets such that $A\subseteq M$ and $\lvert A\rvert=4$ and $B\subseteq X\setminus M$ and $\lvert B\rvert=3$ and $C\cap(A\cup B)=\emptyset$ and $\lvert C\rvert=1$, under the assumption that the map $(A,B,C)\mapsto\begin{cases}(A\cup C,B)&\text{if }C\subseteq M\(A,B\cup C)&\text{otherwise}\end{cases}$ is bijective. This map is not injective. – Sassatelli Giulio Oct 15 '24 at 06:58
  • Namely, if $M={1,\cdots,8}$ and $X={1,\cdots,16}$, then $({1,2,3,4},{9,10,11},{5})\mapsto({1,2,3,4,5},{9,10,11})$ and $({1,2,3,5},{9,10,11},{4})\mapsto({1,2,3,4,5},{9,10,11})$ – Sassatelli Giulio Oct 15 '24 at 06:58
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    For some reason it is conventional to have a very low level of formality in this stuff as a topic. And then when confusion arises there’s no actual rigour to fall back on (unlike something algebraic where you can say ‘write out the working in full and you’ll see the error explicitly somewhere’). This stuff is for some reason often only taught through word puzzles like this and people are more or less encouraged to just write down the answers rather than work towards them. – SBK Oct 15 '24 at 15:50
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    Well for one thing you are counting first choosing Amy, Betty, Carla, and then choosing Darla, as the same thing as first choosing Amy, Betty, Darla and then choosing Carla as different (and the same for the men). You might be doing something else but that's enough over counting to get something 64 times too big. – fleablood Oct 15 '24 at 16:11
  • I think it's worth noting that the odds of each group in the correct answer aren't equally as likely and that's where the difference comes from. The odds of each of the grouping with an extra man is 5/1732640 and the odds of each grouping with an extra woman is only 4/1732640. – Jason Goemaat Oct 16 '24 at 14:40
  • I think the simplest way to answer "how am I overcounting", is that you've not chosen a team of eight with four or five men, but a team of four men plus three women plus a leader. – user3445853 Oct 17 '24 at 12:25

3 Answers3

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There are two possibilities here: You either select five men and three women or four men and four women, which can be done in $$\binom{16}{5}\binom{8}{3} + \binom{16}{4}\binom{8}{4} = 372,008$$ ways.

By designating four of the men as the four men you select and three of the women as the three women you select and then selecting an additional person from the $17$ remaining people, you count each selection with five men and three women five times, once for each of the $\binom{5}{4}$ ways you could designate four of those five men as the four men you select, and each selection with four men and four women four times, once for each of the $\binom{4}{3}$ ways you could designate three of the four women as the three women you select. The reason for the errors is that the set of additional people is not disjoint from the set of men (when you select five men and three women) or from the set of women (when you select four men and four women).

To illustrate with five men and three women: Suppose the men are Andrew, Bruce, Charles, David, and Edward and the women are Fiona, Gloria, and Harriet. Your method counts this selection five times.

$$\begin{array}{l l l} \text{four men} & \text{three women} & \text{additional person}\\ \hline \text{Andrew, Bruce, Charles, David} & \text{Fiona, Gloria, Harriet} & \text{Edward}\\ \text{Andrew, Bruce, Charles, Edward} & \text{Fiona, Gloria, Harriet} & \text{David}\\ \text{Andrew, Bruce, David, Edward} & \text{Fiona, Gloria, Harriet} & \text{Charles}\\ \text{Andrew, Charles, David, Edward} & \text{Fiona, Gloria, Harriet} & \text{Bruce}\\ \text{Bruce, Charles, David, Edward} & \text{Fiona, Gloria, Harriet} & \text{Andrew}\\ \end{array}$$

To illustrate with four men and four women: Suppose the men are Andrew, Bruce, Charles, and David and the women are Esme, Fiona, Gloria, and Harriet. Your method counts this selection four times.

$$\begin{array}{l l l} \text{four men} & \text{three women} & \text{additional person}\\ \hline \text{Andrew, Bruce, Charles, David} & \text{Esme, Fiona, Gloria} & \text{Harriet}\\ \text{Andrew, Bruce, Charles, David} & \text{Esme, Fiona, Harriet} & \text{Gloria}\\ \text{Andrew, Bruce, Charles, David} & \text{Esme, Gloria, Harriet} & \text{Fiona}\\ \text{Andrew, Bruce, Charles, David} & \text{Fiona, Gloria, Harriet} & \text{Esme}\\ \end{array}$$

Notice that $$\color{red}{\binom{5}{4}}\binom{16}{5}\binom{8}{3} + \color{red}{\binom{4}{3}}\binom{16}{4}\binom{8}{4} = \color{red}{1,732,640}$$

N. F. Taussig
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The $\binom{17}{1}$ is the error.

It implies that after you have chosen 4 men and 3 women you still have 17 people to choose from. You do not.

Edit What I mean by you do not is that these 17 people have already been considered and not chosen as part of the original two groups.

One option is to take the two possibilities: the last person is a man, the last person is a woman.

This makes your calculation $\binom{16}{5}\times \binom{8}{3}+\binom{16}{4}\times\binom{8}{4} =372008$

Red Five
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You’ve counted all possible sets of 3 women. And then also added an extra person that could be a woman. So you end up with say women 1,2,3 in the second factor and then woman 4 to be chosen in the final factor. But separately you’d have 1,2 and 4 in the second factor together with counting woman 3 in the final factor

SBK
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  • what did this add to the other answers @garymyerson – suckling pig Oct 16 '24 at 01:41
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    I assume the comment is directed at me? This was actually the first answer. It’s a bit rushed but I tried to say as directly as possible how the overcounting is happening without giving a full solution to the problem. It’s supposed to be in the spirit of the site to do so. But as you can see, what happens is that a lot of people who know how to solve things just write out more of a full solution and people like reading and upvoting them. – SBK Oct 16 '24 at 06:10
  • No worries...@sbk – suckling pig Oct 16 '24 at 07:56