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If the letters of the word "ASSASSINATION" are arranged at random in a row, then the probability that no two $A^{\prime}$ 's come together is equal to
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The correct answer is:
$\frac{15}{26}$
The number of all possible ways to arrange the letters of word ASSASSINATION is $=\frac{13 !}{3 ! 4 ! 2 ! 2 !}$
Now, the number of ways to arrange the letters $S, S, S, S, I, N, T, I, O, N$ is $\frac{10 !}{4 ! 2 ! 2 !}$
and the three $A^{\prime}$ s we must arrange between the gaps or at the end positions and it can be done in ${ }^{11} C_3$.
So, required probability $=\frac{\frac{10 !}{4 ! 2 ! 2 !} \times{ }^{11} C_3}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$
$$
=\frac{3 ! \times \frac{11 \times 10 \times 9}{3 !}}{13 \times 12 \times 11}=\frac{10 \times 9}{13 \times 12}=\frac{5 \times 3}{13 \times 2}=\frac{15}{26}
$$
Now, the number of ways to arrange the letters $S, S, S, S, I, N, T, I, O, N$ is $\frac{10 !}{4 ! 2 ! 2 !}$
and the three $A^{\prime}$ s we must arrange between the gaps or at the end positions and it can be done in ${ }^{11} C_3$.
So, required probability $=\frac{\frac{10 !}{4 ! 2 ! 2 !} \times{ }^{11} C_3}{\frac{13 !}{3 ! 4 ! 2 ! 2 !}}$
$$
=\frac{3 ! \times \frac{11 \times 10 \times 9}{3 !}}{13 \times 12 \times 11}=\frac{10 \times 9}{13 \times 12}=\frac{5 \times 3}{13 \times 2}=\frac{15}{26}
$$
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