5! - 4! x 2 = 72
5! stands for arranging [S] [U] [CC] [E] [SS].
However, we need to exclude these two cases(3S are together) :
{[S][SS]} [U] [CC] [E] and {[SS][S]} [U] [CC] [E], which has 4!
ways to arrange each.
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A simplified version will help to understand better:
You are to arrange 3As and 1B such that exactly 2As are
together,
no. of ways = 3! ([AA] [A] [B]) - 2! x 2 (excluding [(AA)(A)]
[B] and [(A)(AA)] [B] which have been included in the 3!)
3! - 2! x 2 = 2 ways
The answer is obviously correct as the only possible
arrangements are ABAA and AABA