Problem
COM-B2-M02-P005 Derangements of Four Elements
#5
★★☆☆☆ Level 2 of 5
How many permutations of \(4\) elements have no fixed points?
Use events \(A_i\): element \(i\) is fixed.
By inclusion-exclusion, \(4!-\binom41 3!+\binom42 2!-\binom43 1!+\binom44 0!=24-24+12-4+1=9\).
It is important to explain the term \(\binom4i(4-i)!\).