Consider all the permutations of the word ASSASSIN, then the probability that

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No two alike letters are together is $\frac{41}{140}$

Exactly one pair of S are together is $\frac{2}{7}$

First A occurs before second S is $\frac{3}{5}$

Second A occur after third S is $\frac{3}{5}$

answer is C, D.

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Detailed Solution

ASSASSIN
Permutation when no 2S are together minus
permutation when no 2S are together but 2A’s
are together. $\Rightarrow \frac{^{5} C_{4} \times \frac{4!}{2!} -^{4} C_{4} \times 3!}{\frac{8!}{4! 2!}} = \frac{9}{140}$
Consider A and S to be alike so no. of arrangement $= \frac{8!}{6!}$
If A comes at first place $A \underset{}{\underset{⎵}{- - - - - - -}} \frac{5!}{4!}$
If S comes at first place $S A \underset{}{\underset{⎵}{- - - - - - -}} \frac{4!}{3!}$
Hence first A comes before second S is $(\frac{8!}{6!} \frac{[\frac{5!}{4!} + \frac{4!}{3!}]}{\frac{8!}{4! 2!}}) = \frac{3}{5}$