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$30$ people want to play capture the flag. There are two teams. Each team has ten people. How many ways to choose the teams? I figure there are $30\choose 10$$20\choose 10$ ways to choose members for the team. I am told that this is "double counting", and the answer should be half of $30\choose 10$$20\choose 10$Why is this so?

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  • $\begingroup$ Are there two teams or three? $\endgroup$ – N. F. Taussig Sep 14 '17 at 0:14
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    $\begingroup$ Choosing the first ten to be on one team and the second ten to be on the other team is the same as choosing the second ten to be on one team and the first ten to be on the other. $\endgroup$ – lulu Sep 14 '17 at 0:17
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    $\begingroup$ Alternatively, avoiding the "division" issue, first pick which ten people are waiting on the bench and not playing. Then among the twenty people who are going to be playing, one of them will be the youngest. Pick who the nine people are who will be on his team. This gives an answer of $\binom{30}{10}\binom{19}{9}$. $\endgroup$ – JMoravitz Sep 14 '17 at 0:22
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    $\begingroup$ Sometimes I find it easier to reduce the counting required to make it easier to "see" an answer. You have three people and you must pick 2 teams, one person per team (it's a chess tournament). If you pick person number 1 for the first team, and person 2 for the second team, that's the same as picking 2 and then 1, because it's still the same two people facing off against each other. Makes sense? Hence that's "double counting." (It's been ages since I've done these, let's see if I get corrected.) $\endgroup$ – markspace Sep 14 '17 at 0:43
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The question is not 100% clearly phrased. You could also consider your answer correct.

Say there is a red and a blue team. Then you correctly figured that there are $\binom{30}{10}\binom{20}{10}$ ways to pick 10 players for 10 red and 10 players for the blue team.

But now you can argue that swapping the team colors does not change anything. So if you do not care which team is red and which is blue but only which players play together the number of possibilities is reduced in half.

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Actually, there are two types of situations

  • The teams are labelled (distinguishable), e.g. Lions and Tigers
    Here you would not divide by $2$,
    as being on the Lions team is not the same as being on the Tigers team.

  • The teams are unlabelled (indistinguishable)
    Here you would divide by $2$, as explained in the various comments.

In the absence of any explicit mention of teams being labelled, we presume that they are unlabelled.
As a safeguard, you could add the above remark to your answer, if remarks are permitted.

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  • $\begingroup$ @wonce: I can't understand why your answer wasn't showing when I was typing mine. Anyway, I'll let it stand with credit to you ! $\endgroup$ – true blue anil Sep 14 '17 at 7:26

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