# Combinatorics. How many ways can you split 8 people into pairs for the game of bridge.

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?

• 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 $N = 8$ ? – Dominik Kutek Sep 25 at 19:21
• 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 ... – geguze Sep 25 at 19:32

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 $$7!! = 7\times 5\times 3\times 1$$