We have 20 chess players. They each find opponents and start playing. How many possibilities are there for how they are matched up, assuming that in each game it does matter who has the white pieces?
Attempt
Note the first player can play against 19 other opponents so that we 19 choices for first player. Same goes for the second one so that is 19 possibilities again and so on. In total then we have $19^{20}$ possible matches. The answer tho is given to be $\frac{ 20!}{10!}$ but I really cannot see why this is so.
What they are saying in this. Line the $20$ players up in a row. There are $20!$ ways to do this. Now put the first two players at the first table, the next pair of players at the second table, and so on. We can also suppose that the odd-numbered players in the line get white and the even numbered players get black, so that having A,B as the first two player is different from having B,A. But it does not matter whether A and B play at the first table or the fifth, so we have to divide by the $10!$ ways of shuffling the pairs.
Therefore, the answer is $$\boxed{\frac{20!}{10!}}$$