Exercise 1.6

Question

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 (in a chess game, one player has the white pieces and the other player has the black pieces)?

fifthist x · Accepted Answer

Line up the 20 players in some order then say the first two are a pair, the
next two are a pair, etc. This overcounts by a factor of 10! because we don't 
care about the order of the games. So in total, we have
    $$ \frac{20!}{10!} $$
ways for them to play. This correctly counts for whether player A plays 
white or black. If we didn't care we would need to divide by 2$^{10}$.

Another way to look at it is to choose the 10 players who will play white then 
let each of them choose their opponent from the other 10 players. This gives a 
total of

$$ {20 \choose 10} 	imes 10!$$

possibilities of how they are matched up. We don't care about the order of the players who play white but once we've chosen them the order of the players who play black matters since different orders mean different pairings.

Exercise 1.6

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Exercise 1.6

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