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?
Combinatorics. How many ways can you split 8 people into pairs for the game of bridge.
combinatoricspermutations
Best Answer
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 $7!! = 7\times 5\times 3\times 1$