1 2 3 4 5 6 7 8 9 10 11 12

hm

This is harder than it seems.

You can swap 2 adjacent seats once.

For example, I can swap 1 and 2.

2 1 3 4 5 6 7 8 9 10 11 12

However, I can't swap a number twice.

So once I swap 1 and 2, I can't then swap 1 and 3.

For 6 swaps, there is 1 way you can do this.

Swap pairs (1, 2), (3, 4), ...(11, 12)

For 5 swaps there is 6 ways of doing this.

Swap pairs (3, 4), (5, 6)... (11, 12) not swapping (1, 2)

Swap pairs (1, 2), (5, 6)... (11, 12) not swapping (3, 4)

etc.

I think you just have to count the number of ways you can swap, from 6 swaps to 0 swaps.

=^._.^=