Badminton Game : Permutations and Combinations

Question

If 40 players are participating in a single's badmintom tournament, then number of all possible pairings for the first round is?

My way of thinking is:

Out of 40 players, I can select any two in 40C2 ways. From the remaining 38 players, in 38C2 ways and continuing like this, until I ultimately get 2C2.

So my final answer is: 40C2 * 38C2 * 36C2.... 4C2 * 2C2 which on solving brings : $$\frac{40!}{(2!)^{20}}$$.

But the final answer is: $$\frac{40!}{(2!)^{20}(20!)}$$.

Why is the 20! there? What does it signify?

Answer 1

You can think of there being $40!$ ways to put the players in line, then you take the first two for the first game, the next two for the next game, and so on. Each pair can come in either order, so you divide by $2^{20}$. Then the pairs can come in $20!$ different orders, so you divide by $20!$. If you care about the order the pairs are selected, say because the first pair in line plays the first game, you do not divide by $20!$.

Answer 2

The organizers count ABC as the same draw as ACB, BAC, BCA, CAB and CBA.

Your answer would count all six.

There are $20$ such pairings, and so $20!$ 'repeats', so they have divided by $20!$.

The last $2!$ identifies a drawing AB to a drawing of BA.