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My Trial:: First we will count Total no. of arrangement of words without restriction.

which is $\displaystyle = \frac{7!}{3!\times 2!} = 420$

Now Total no. of arrangement in which two $\bf{C}$ are together

which is $\displaystyle = \frac{6!}{3!} = 120$

Now Total no. of words in which no two $\bf{S}$ are togrther, is $\displaystyle = \binom{5}{3}\times \frac{4!}{2!} = 120$

Now I did understand How can i Calculate after that

So Help please

Thanks