Given 11 chess players and 5 distinct tables, in how many ways can we pair them to play (color does matter)?
My problem is that I have found two approaches, both of which give different numbers, and I am not sure what is missing in one or double-counted in the other.
The first approach is to just take any permutation of the $11$ players, as the tables are distinct there are 11 unique spots (one of them is not playing), so we can just place the players according to the permutation. This gives $11!=39916800$ possible games.
The second approach is to first choose pairs, the first player has 10 choices, the next has 8, … and then we multiply by $2^5$ to account for the colors of each pair, and finally multiply by the number of ways to seat them at tables (so multiply by $5!$), giving
$$10\cdot8\cdot6\cdot4\cdot2\cdot2^5\cdot120=14745600$$
My intuition is that the first approach is correct, but then I am not sure which pairings are missing in the second one..
Best Answer
When choosing pairs in the second approach, note that there are ${ 11 \choose 2}$ ways to choose the pair to play at the first (distinct table). As such, this method yields:
$$ { 11 \choose 2} { 9 \choose 2 } { 7 \choose 2} { 5 \choose 2 } { 3 \choose 2 } { 1 \choose 2 } \times 2^5. $$
By inspection, this is equal to $11!$.