Question

In how many ways can the letter of the word SUCCESS be arranged so that
(i) The two $C^{\prime }s$ are together bot no two $S^{\prime }s$ are together.
(ii) No two $C^{\prime }s$ and no two $S^{\prime }s$ are together.

Answer 1

Take out three $S^{\prime }s$ and tie up two $C^{\prime }s$ so that we are left with U,E and one bundle of $C^{\prime }s$. Thus we have 3 letters which can be arranged in $3!\left(\frac{2!}{2!}\right)=6$ways.
We have 4 places (two in between and two in the
corners). In these 4 places 3 different objects can be arranged in ${}^4{P}_3$ ways but three alike can be arranged in $\frac{1}{3!}{.}^4{P}_3$ ways i.e. 4 ways.
Hence by fundamental theorem, the number of desired arrangements is $4\times 6=24$.
(ii) Now take out $SSS$ and arrange remaining 4 letters out of which two are alike $\therefore \frac{4!}{2!}=12$. There will be 5 gaps in between these 4 (3 in between 2 at corners) in which three alike $SSS$$ can be placed in
$\frac{{}^5{P}_3}{3!}$ways, i.e. $\frac{5.4.3}{6}=10$ ways.
Hence by fundamental theorem there will be $12\times 10=120$ ways in which all the $S^{\prime }s$ are separated. In the above 120 ways no two $S^{\prime }s$ are together. Also by part (i) no two $S^{\prime }s$ are together but two $C^{\prime }s$ are together in 24 ways.
$\therefore$ In $120-24=96$ ways no two $S^{\prime }s$ and no two $C^{\prime }s$ are together.