How many permutations can be made of word "ASSASSIN" such that only $2$ Ss are together? How many permutations can be made of word "ASSASSIN" such that only $2$ Ss are together?
I have been doing by taking $2$ S together in a group and arranging them as
$$\frac{7!}{2!2!}=1260$$
which is greater than total number of permutations of the word i.e. $840$.
 A: In a permutation of the word "ASSASSIN", consider the number of letters $x_i\geq 0$ between the $i$-th $S$ and the previous one:
$$\underbrace{\dots}_{x_1}S
\underbrace{\dots}_{x_2}S
\underbrace{\dots}_{x_3}S
\underbrace{\dots}_{x_4}S
\underbrace{\dots}_{x_5}$$
Then 
$$x_1+x_2+x_3+x_4+x_5=4$$ 
and such spaces can be filled with the letters $A$, $A$, $I$, $N$, in $4!/2=12$ ways.
In order to have exactly two $S$ together we have $3$ distinct cases:
i) $x_2=0$, $x_3\geq 1$ and $x_4\geq 1$;
ii) $x_3=0$, $x_2\geq 1$ and $x_4\geq 1$; 
iii) $x_4=0$, $x_2\geq 1$ and $x_3\geq 1$.
Hence, by the stars and bars method,  the required number of ways is
$$12\cdot 3\cdot \binom{(5-1)-2+3}{3}=360.$$
A: First, arrange the letters $A, A, I, N$.  There are $\binom{4}{2}$ ways to choose the positions of the $A$s and $2!$ ways to arrange the remaining letters.  Doing so creates five spaces in which the $S$s may be placed, three between successive letters and two at the ends of the row.  For instance, if the arrangement is $AAIN$, we have
$$\square A \square A \square I \square N \square$$
We have five choices where to place the double $S$.  Once we choose its position, there are four empty spaces left in which to place the remaining $S$s.  In order to separate them, we must choose two of these four spaces, which can be done in $\binom{4}{2}$ ways.  Hence, the number of admissible arrangements is
$$\binom{4}{2}2!\binom{5}{1}\binom{4}{2} = 360$$
A: Take two $S$'s together and pretend this is a new character. The number of ways to divide this character and the two remaining $S$'s over the seven positions equals:
$${3 \choose 1} {2 + 4 - 1 \choose 4 - 1} = 3 \cdot 10 = 30$$
Indeed, there are three ways to sort them, and there must be at least one empty space between each of them (which leaves two spaces to divide among four possible locations, a stars and bars problem). We then have to divide the remaining characters, and the total number of combinations thus equals:
$$30 \cdot {4 \choose 2, 1, 1} = 30 \cdot 12 = 360$$
