There are $4$ consonants and $1$ group of vowels, so there are $5$ elements to permute. Yes, this is a combinatorial problem because we are counting the number of possibilities that satisfy certain conditions.
With the inclusion-exclusion principle, I would like to compute the following:
(Total number of permutations) - (Number of permutations with $2$ $C$'s together) - (Number of permutations with $2$ $S$'s together) + (Number of permutations with $2$ $C$'s together and $2$ $S$'s together) + (Number of permutations with $3$ $S$'s together) - (Number of permutations with $2$ $C$'s together and $3$ $S$'s together).
Total number of permutations: $\frac{7!}{3!2!}$. There are $7!$ orderings, but we divide out by the repeated letters.
Number of permutations with $2$ $C$'s: $\frac{6!}{3!}$. There are $6!$ orderings when we treat $CC$ as a single character, but we divide out by the repeated $S$'s.
Number of permutations with $S$ $S$'s: $\frac{6!}{2!}$. Put $2$ of the $S$'s together and leave the third one out of the pair. Divide out by the repeated $C$'s.
Number of permutations with $2$ $C$'s and $2$ $S$'s: $5!$. Put the $2$ $C$'s and $2$ of the $S$'s as a single letter.
Number of permutations with $3$ $S$'s: $\frac{5!}{2!}$. Put all $3$ of the $S$'s together and divide out by the $C$'s.
Number of permutations with $2$ $C$'s and $3$ $S$'s: $4!$. Put the $2$ $C$'s and $3$ $S$'s as single letters.
Best Answer
These problems quickly get out of hand if the words are long and there are lots of multiple letters. Here is a sophisticated solution that uses ideas from algebraic combinatorics. I learned it from Jair Taylor's wonderful answer here. See this question also.
Define polynomials for $k\geq 1$ by $q_k(x) = \sum_{i=1}^k \frac{(-1)^{i-k}}{i!} {k-1 \choose i-1}x^i$. Here are the first few polynomials: $$q_1(x)=x,\quad q_2(x)=x^2/2-x,\quad q_3(x)=x^3/6-x^2+x.$$
The number of permutations with no equal neighbors, using an alphabet with frequencies $k_1,k_2,\dots$ is:
For the "success" problem, the product of the $q$ functions is $$ q_3(x)\, q_2(x)\, q_1(x)^2=(x^3/6-x^2+x)(x^2/2-x)x^2 = x^7/12-2x^6/3+3x^5/2-x^4,$$
and performing the integral gives the answer 96.