pairing possibilities in chess game There are 20 people at a chess club on a certain day. 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 (and who has the black ones). 
I thought it might be $$\large2^{\frac{20(20-1)}2}$$ is this correct?
 A: A set of twenty distinct items can be ordered in $20!$ ways.
A set of $10$ distinct pairs can be ordered in $10!$ ways.
A pair of distinct items can be ordered in $2!$ ways.
Since we don't care about the order of the pairs, and only care about the the order within each pair, then a set of $20$ distinct items can be subdivided into an unordered set of $10$ distinct ordered pairs in $\frac{20!}{10!}$ ways.

If it doesn't matter about the order within each pair then a set of $20$ distinct items can be subdivided into an unordered set of $10$ distinct unordered pairs in $\frac{20!}{10!\;2!^{10}}$ ways.
A: If we arrange the players in some order, say of height, and let the shortest player choose first, then there are 19 people for him to choose from. Then let the next shortest remaining player choose, and there are 17 people left to choose, and so on. Thus the number of ways to choose pairs is
$$
19\times17\times15\times13\times ...\times3\times1
$$
This is equal to 
$$
{20!\over 2^{10}10!}
$$
If we care about which is white and which is black, once we have chosen the pairs, the number of ways to choose who is white and who is black among the ten pairs is $2^{10}$, so we get
$$
{20!\over2^{10}10!}\times2^{10}={20!\over10!}
$$
A: I would like to share how I ended up thinking about this.
To come up with a single instance of pairings first choose who gets to play white. There are $20\choose{10}$ ways to pick who gets to play white. Then permute the remaining 10 players, and pair them up with the white players by order (or index).
Example:
Say we randomly chose 1 3 5 7 9 11 13 15 17 19 to play white.
Then we permuted the rest and got 20 18 16 14 12 10 8 6 4 2.
Then we have the pairings:
{(1,20), (3,18), (5,16), (7,14), (9,12), (11,10), (13,8), (15,6), (17,4), (19,2)}
So we end up having $20\choose{10}$ = $\frac{20!}{10!10!}$ ways to choose who plays white, and for each those we will have $10!$ permutations of who to pair them with, this makes it so the number of pairs is $\frac{20!}{10!10!}10! = \frac{20!}{10!}$.
A: Here we have 2 things to keep in mind:


*

*Number of ways of matching up

*Arrangements does matter


Firstly, Number of ways of matching up players:
1st player: 19 player options; chooses: 1; players left after choosing: 18
2nd player: 17 player options; chooses: 1; players left after choosing: 16
.. continuing this way, we end up with,
10th player: 1 player options; chooses: 1; players left after choosing: 0
hence, number of ways to match up =  
$$
19 * 17 * 15 * ... 3*1
= \frac{20!}{2^{10}*10!}
$$
Now, considering arrangements, for each pair of players, there are 2 possibilities of choosing white/black.
So, for 10 pair of players, the number of arrangements is 
$$
{2^{10}}
$$
So, number of ways to match up players, considering arrangements of black and white sides is
$$
2^{10}* \frac{20!}{2^{10}*10!}
 = \frac{20!}{10!}
$$ 
A: First we choose 2 from 20 to form pair 1, then choose 2 from 18 to form pair 2... until last 2 players to form pair 10: there are totally $\binom{20}{2}\binom{18}{2}\ldots\binom{2}{2}$ possible combinations to form 10 ordered pairs.
However, in this problem, order of pairs does not matter. Since there are 10! ways to arrange 10 pairs, we should divide it by 10!.
Also, in each pair, either player can play black or white, so we should multiply 2 for each pair, that's $2^{10}$.
Put them together, we have:
$$\frac{\binom{20}{2}\binom{18}{2}\ldots\binom{2}{2}}{10!}\times2^{10} = \frac{20!}{10!}$$
