If the letters of the word ASSASSINATION are arranged at random.Find the probability that no two A's are coming together

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

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.