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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..

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The first approach perfectly makes sense.

In the second approach, when we say first player $p$ has $10$ options, we explicitly choose a player and choose it's opponents from $10$ options. But then we skip the case where $p$ does not play at all. This shows there are some cases that are not counted but still does not cover all the cases that are not counted. Missing cases probably come from other assumptions like second chosen player having $8$ options, third chosen player having $6$ options, etc.

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    $\begingroup$ Ohh yeah! That makes perfect sense. Not sure how I missed that... Thank you! $\endgroup$ – Nescio Apr 24 at 10:00
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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 $11!$.

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