Solution: 2018 Fall Final - 7

First, we take a E and a U and treat them as a single entity, so the conditions are met

Since the entity can be either EU or UE, there are double the possibilities: 2

We now act as if we only have 6 entitieS.

There are 6 possible positions for the EU/UE to go: 6

For the 3 $S$‘s, we choose 3 positions out of the 5: $ \binom{5}{3} $

For the 2 $C$‘s, we choose 2 positions out of the 2: $ \binom{2}{2} = 1 $

$ 2 \cdot 6 \cdot \binom{5}{3} $

$ = 2 \cdot 6 \cdot \frac{5!}{3!2!} $

$ = 2 \cdot 6 \cdot \frac{5 \cdot 4}{2} $

$ = 2 \cdot 6 \cdot 10 $

$ = 2 \cdot 60 $

$ = 120 $