Number of possible chess pairs where order and position matter

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 $11!$.

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.