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

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Did you intend to write the word "Success" with that exact capitalization? What I mean is, are all of the s's to be considered interchangeable? –  Zev Chonoles Jun 24 '12 at 13:08
    
No, ignore the capitalization. –  barricade Jun 24 '12 at 13:11

6 Answers 6

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.

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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
$${5\choose 3}{4!\over 2!}-{4\choose 3}{3!}=120-24=96. \ \ \ \ $$

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120 - 24* = 96. –  Ishaan Singh Jun 24 '12 at 16:56
    
@IshaanSingh Thanks, I've corrected it. –  Byron Schmuland Jun 24 '12 at 17:23

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]);
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Dunno what's wrong with my logic but I got 72. Got $\frac{6!}{3!} $,$\frac{6!}{2!} - \frac{5!2!}{2!}$ and $5! - 4!2!$ –  Dan Jun 24 '12 at 18:07
    
why did you multiply by 2! ?? –  Santosh Linkha Jun 24 '12 at 18:10
    
for e.g to find at least two s together I permute s,u,c,c,e,ss then I subtract the number of permutations with s and ss are together. and the 2! means I'm accounting for the permutation of s and ss when they are counted as a single object... –  Dan Jun 24 '12 at 18:34
    
Let there be $n$ elements, $t$ being identical and rest unique. The no of arrangements of $t-1$ consecutive identical elements, for some strange reasons seems to be$(n-(t-1))! - (n-t)!$ –  Santosh Linkha Jun 24 '12 at 18:44
    
@Dan try for $|a|b|s|s|$ and $|a|s|s|s|$ the cases where you get consecutive $ss$ –  Santosh Linkha Jun 24 '12 at 18:52

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.)

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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

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You did not factor in the condition that the 's' can not repeat next to each other, nor the 'c'. –  Willie Wong Apr 22 '13 at 13:07

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.

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1  
This is incorrect, as Willie Wong's comment pointed out on said answer. –  user61527 Nov 9 '13 at 1:52

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