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I'm really bad at counting problems, so I'm not surprised I couldn't get this one.

The Sagebrush student council has 6 boys and 6 girls as class representatives. Two different subcommittees, each consisting of 2 boys and 2 girls are to be created. If no student can be in both subcommittees, how many different combinations of subcommittees are possible?

So I first find the number of ways to make the first subcommittee, which I think is $2\cdot(6\operatorname{C}2)=30$(6 choose 2 because we are choosing 2 people out of 6 original; times 2 for both girl and boy choices). For the second subcommittee, I have $2\cdot(4\operatorname{C}2)=12$(4 choose two because we are choosing two people out of the now limited 4; times 2 for both girl and boy choices). Adding them together, I have a total of $42$. This is wrong.

What did I do wrong? How should I solve this problem?

Thanks! Your help is appreciated!

Max0815

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    $\begingroup$ Are the subcommittees considered distinct or indistinct? Does the first subcommittee work on a different issue than the second subcommittee? By that, I mean if you have the first two boys and first two girls on the first subcommittee and the last two boys and last two girls on the second subcommittee, is this considered a "different" outcome than having the last two boys and last two girls on the first subcommittee with the first two of each in the second subcommittee? $\endgroup$ – JMoravitz Mar 22 at 1:17
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    $\begingroup$ You have multiplied by two in several of these situations. Why? If you have $\binom{6}{2}$ choices for who the boys are on the "first" subcommittee, and you have $\binom{6}{2}$ choices for the girls on the first subcommittee... you multiply these together to get $\binom{6}{2}\times \binom{6}{2}$, not $2\times \binom{6}{2}$. $\endgroup$ – JMoravitz Mar 22 at 1:18
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Hint:

Since for the first subcommittee you are choosing boys and choosing girls, the number of ways of doing that has to be multiplied: $$\binom{6}{2} \times \binom{6}{2} \ .$$

The second subcommittee follows the same reasoning, but now we have $4$ boys and girls left.

As for the total number of combinations of the subcommittees, we will need to have the first subcommittee and the second subcommittee.

Can you compute everything now?

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  • $\begingroup$ Ah, I did. Thank you very much! $\endgroup$ – Max0815 Mar 22 at 1:31
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Since I am trying to choose two subcommittees and they are not independent of each other, I have to multiply all of my numbers together to get the answer. There are $(6\operatorname{C}2)^2$ ways to choose the first subcommittee, and $(4\operatorname{C}2)^2$ ways to make the second subcommittee, so thus, the total number of ways to make 2 subcommittees is $(6\operatorname{C}2)^2\cdot (4\operatorname{C}2)^2=8100$.

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