Question #40586

This question tests the 'Set Complement Boundary' trap within a restricted universe.

Students commonly assume that fixing one element in a derangement reduces the problem to a standard derangement of size $n-1$ (i.e., incorrectly assuming $|D \cap C_1| = D_4 = 9$).

However, because the fixed element $\pi(1)=2$ is not mapped to its own index, the element $1$ is freed from any positional restriction in the remaining mapping.

This shifts the boundary of the forbidden set, requiring a custom inclusion-exclusion calculation for the sub-universe ($4! - 3 \times 3! + 3 \times 2! - 1! = 11 \neq 9$).

Failing to recalculate these shifted boundaries leads directly to the trap answer of 22.