Arrangement of the word 'Success'

Number of ways the word 'Success' can be arranged, such that no two S's and C's are together.

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:

$$\int_0^\infty \prod_j q_{k_j}(x)\, e^{-x}\,dx.$$

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.

We start with all arrangements with non-consecutive "S"s, then subtract those where the "C"s are together.

That is, we begin with the arrangements with non-consecutive "S"s over the alphabet {S,U,C,C,E,S,S} and then subtract the arrangements with non-consecutive "S"s over the alphabet {S,U,CC,E,S,S}. Note the double "C" in the second alphabet.

Using the formula from my answer here, we get
$${5\choose 3}{4!\over 2!}-{4\choose 3}{3!}=120-24=96. \ \ \ \ $$