Question

The probability that the $3N^{\prime }$s come consecutively in the arrangement of the letters of the word ${}^{\prime }CONSTANTINOPL{E}^{\prime }.$

• A. $\frac{3}{91}$
• B. $\frac{6}{91}$
• C. $\frac{9}{91}$
• D. $\frac{2}{91}$

Answer 1

The correct option is A $\frac{3}{91}$

The total number of arrangement of the letters of the word ${}^{\prime }CONSTANTINOPL{E}^{\prime }$ is $\frac{\left(14\right)!}{3!2!2!}$

Since $3N^{\prime }$s are consecutive, then considering all the $3N^{\prime }$s as single letter, the total favourable number of arrangements is $\frac{\left(12\right)!}{2!2!}$
Therefore, required probability is
$\frac{\frac{\left(12\right)!}{2!2!}}{\frac{\left(14\right)!}{3!2!2!}}=\frac{3!}{14\times 13}=\frac{3}{91}$