Problem
COM-B1-M02-P021 Derangements of Five
#21
★★★★☆ Level 4 of 5
How many permutations of \(1,2,3,4,5\) leave no number in its original position?
Apply inclusion-exclusion over fixed points.
Total: \(5!=120\). By inclusion-exclusion over specified fixed points: \(120-5\cdot4!+10\cdot3!-10\cdot2!+5\cdot1!-1=120-120+60-20+5-1=44\).
First serious derangement.