Mathematics

This is a derangement-adjacent counting problem where you need bijections with a forbidden assignment. The key insight is to use complementary counting: find all bijections from $A$ to $B$, then subtract those where $f(a_1) = b_1$. When $f(a_1) = b_1$, the remaining three elements must form a bijection between the remaining sets—a simpler subproblem. This transforms a constraint into a straightforward subtraction.

This is a derangement-adjacent counting problem where you need bijections with a forbidden assignment. The key insight is to use complementary counting: find all bijections from $A$ to $B$, then subtract those where $f(a_1) = b_1$. When $f(a_1) = b_1$, the remaining three elements must form a bijection between the remaining sets—a simpler subproblem. This transforms a constraint into a straightforward subtraction.