Number of ways to arrange a tournament of 8 people

Question

If you are arranging a 1vs1 tournament of 8 people, how many different ways can you arrange the first round? A set of pairings is different if someone has a different opponent. For example, you can arrange a tournament of 4 as follows: $\{AvB, CvD\}$, $\{AvC, BvD\}$, ${AvD, BvC\$} = 3 different ways this can be arranged.

Answer 1

The first round of the tournament can be arranged in $\frac{8!}{2^4}=2520$ ways

The first contestant can be chosen in $8$ ways, for each of those choices, there are $7$ choices for the second contestent of the pair. But this way, we have counted all pairs twice, because we have counted the pair $(AB)$ as the contestant $A$ first and the pair $(BA)$ as the contestant $A$ second in the pair. So the total number of choices for the first pair is $\frac{8\cdot 7}{2}=28$.

A more general result states that from $n$ elements you can take $\binom{n}{2}=\frac{n!}{2!(n-1)!}=\frac{n(n-1)}{2}$ subsets of two elements. ie there are $\frac{n(n-1)}{2}$ pairs.

If we have $n=2m$ contestants total, I claim there are $\frac{n!}{2^m}$ ways to organise the first round. The proof is by induction. For $n=2$ it is clear. The first pair can be chosen in $\frac{n(n-1)}{2}$ ways. For each of those choices, we are left with $n-2=2(m-1)$ contestants, by induction the rest of the pairs can be organised in $\frac{(n-2)!}{2^{m-1}}$ ways. This means the total number of possible arrangements for the first round are $\frac{n(n-1)}{2}\frac{(n-2)!}{2^{m-1}}=\frac{n!}{2^m}$

If the number of contestants were odd, $n=2m+1$, you can aply similar arguements and find that again there are $\frac{n!}{2^m}$ ways to roganise the first round

Aplying that for $n=8=2\cdot 4$ gives for the first round $\frac{8!}{2^4}=\frac{40320}{16}=2520$ possible ways.

Edit: Corrected a BIG mistake I did and gave a general formula