Problem
COM-B1-M02-P024 Five Pairs of Letters Without Adjacency
#24
★★★★★ Level 5 of 5
How many permutations of \(A,A,B,B,C,C,D,D,E,E\) have no two identical adjacent letters?
Use inclusion-exclusion over the events \(AA\), \(BB\), \(CC\), \(DD\), \(EE\) are adjacent.
There are \(10!/(2!^5)=113400\) total permutations. If \(k\) letters are chosen whose pairs are glued, we get \(C(5,k)(10-k)!/(2!^{5-k})\) arrangements. Inclusion-exclusion gives \(113400-113400+50400-12600+1800-120=39480\).
Strong problem: the same idea with more layers.