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.
 A: 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.
A: 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
$${5\choose 3}{4!\over 2!}-{4\choose 3}{3!}=120-24=96. \ \ \ \ $$
A: The total number of permutation of letters (T)= $\frac{7!}{2!3!}$
With two cc together (A)= $\frac{6!}{2!}$
With three ss together (B)=  $\frac{6!}{2!} - \frac{5!}{2!}$
With both ss and cc together (C)= $5! - 4!$
Answer = T - A - B + C = 96
EDIT::
The number of unique permutation of consecutive $t-1$ $s$'es out of $t$ $s$'es in total of $n$ elements is given by $(n-(t-1))! - (n -t)!$ and this does not include $t$ s'es. 
Matlab code for answer:
P = unique(perms(['s' 'u' 'c' 'c' 'e' 's' 's']), 'rows');
count = 0;
for i = 1:length(P)
    for j = 1:6
        if P(i,j) == P(i,j+1)
            count = count+1;
            break;
        end
    end
end
disp([length(P), count, length(P)-count]);

A: Place the letters S separated by some space. This gives four possible spaces for the remaining letters including the ends. Choose 2 of these spaces for the C letters; that is 6 possibilities. If the C letters are placed at the ends are two ways to place the U and E.
The configuration would look like CSXSXSC. There are thus 2 possibilities. If only one C is at an end, the other is interior say CSCSXS then we can place the U&E in the position of the X and then place the other letter in one of 7 positions for a total of 14 possibilities.
If both C letters are interior then we have SCSCS and so we can place the U in any of 6 positions and then the E in any of 7 positions so we get 42 possibilities. This gives a grand total of 42+4*14+2=100. (I've probably overlooked something.)
A: In my opinion the answer would be as follows
$$
\frac{7!}{3!\cdot 2!} = 420
$$ ways we can arrange word success
by: Javed Masood - FUUAST
A: The total number of permutation of letters (T)= 7! 3!2!   
Answer = N!/(N1!*NK!) = 7!/(3!*2!)= 420  
The answer from April 22 at 0535 is correct.  the rest are dead wrong.  
