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The title explain the problem. First I would choose two random people out of 8, there is 4 ways to choose two people, then there's still 6 people left. Then again two people should be chosen, so there are 3 ways to do it. Then only 4 people are left. Again we choose two random people from those who have been left: 2 ways to arrange them into pair. Lastly there are only two people left, so there is only $1$ way to choose a pair. So together it will make $4\cdot 3 \cdot 2=24.$ I would like to know whether my answer is correct and also I notice that the result is $\left(\frac {N}{2} \right)!$ so what's the reasoning behind this?

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    $\begingroup$ Wait, you want to count ways to split 8 people into pairs for the game of bridge and you're getting number between $0$ and $1$? And is $N = 8$ ? $\endgroup$ – Dominik Kutek Sep 25 at 19:21
  • $\begingroup$ Let's think: if you have 2 people you only have one option. If you had three people ${A,B,C}$ then you can choose $(A,B)$ or $(A,C)$ or $(B,C)$ so three. Clearly, it must be an integer number ... Check "n choose k", you can find plenty of information about it ... $\endgroup$ – geguze Sep 25 at 19:32
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Assuming you do not "name the teams" or order the teams, etc...

Without loss of generality, assume that everyone's age is different.

Choose the partner for the youngest person: $7$ options

From those remaining people (not the previously selected pair), choose the partner for the youngest remaining: $5$ options

Continue in this fashion, giving a final total of $7!! = 7\times 5\times 3\times 1$

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