COM-B2-M02-P016 Forbidden Positions in Rook Form

The forbidden cells are \((1,1),(1,2),(2,1)\). Track which pairs of restrictions are compatible.

There are \(5!=120\) permutations. One violation fixes one cell and gives \(4!\) permutations; there are \(3\) such violations, so subtract \(72\). Two violations are compatible only for the pair \((1,2)\) and \((2,1)\). The pair \((1,1),(1,2)\) is impossible because \(\pi(1)\) cannot be both \(1\) and \(2\). The pair \((1,1),(2,1)\) is impossible because two values cannot both land in \(1\). The compatible pair gives \(3!\) permutations. A triple violation is impossible. The answer is \(120-3\cdot24+6=54\).

Shows that intersections do not always have the same size.