Question

If the letters of the word MATHEMATICS are arranged arbitrarily, the probability that C comes before E, E before H,H before I and I before S is

• A. $\frac{1}{75}$
• B. $\frac{1}{24}$
• C. $\frac{1}{120}$
• D. $\frac{1}{720}$

Answer 1

The correct option is C $\frac{1}{120}$

The total numbers of arrangements is $\frac{11!}{2!2!2!}=\frac{11!}{8}$
The number of arrangements in which C, E, H, I, S appear in that order = $\left(\begin{array}{c}11\\ 5\end{array}\right)\frac{6!}{2!2!2}=\frac{1}{8.5!}$
Probability =$\frac{11!}{8.5!}÷\frac{11!}{8!}=\frac{1}{5!}=\frac{1}{120}$