Problem
COM-B1-M02-P011 Derangements of Four
#11
★★☆☆☆ Level 2 of 5
How many permutations of \(1,2,3,4\) leave no number in its original position?
Use inclusion-exclusion over fixed points.
Total: \(24\). With one specified fixed point: \(3!\), and there are \(4\) choices. With two: \(2!\), and \(6\) pairs. With three: \(1!\), and \(4\) choices. With four: \(1\). Answer: \(24-4\cdot6+6\cdot2-4\cdot1+1=9\).
Gentle entry into inclusion-exclusion.