$\begin{array}{1 1}(A)\;\large\frac{15}{26}\\(B)\;\large\frac{5}{26}\\(C)\;\large\frac{25}{26}\\(D)\;\large\frac{9}{26}\end{array} $

Given the word : ASSASSINATION

No two A's are coming together

ASSASSINATION can be arranged in $\large\frac{13!}{3!\times 4!\times 2!\times 2!}$ ways

No two A's coming together XSXSXSXSXIXNXTXIXOXNX

A can be put in place of any X.It can be put in 11 places in $11C_3$ ways

Number of favorable conditions =$\large\frac{10!}{(4!\times 2!\times 2!\times 11C_3)}$

Step 2:

$\therefore$ Probability =$\large\frac{\Large\frac{10!}{(4!\times 2!\times 2!\times 11C_3)}}{\Large\frac{13!}{(3!\times 4!\times 2!\times 2!)}}$

$\Rightarrow \large\frac{15}{26}$

Hence (A) is the correct answer.

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