First slide

Theorems of probability

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

Moderate

A

$\frac{1}{75}$
B

$\frac{1}{24}$
C

$\frac{1}{120}$
D

$\frac{1}{720}$

Solution

$The total numbers of arrangements is \frac{11!}{2! 2! 2!} = \frac{11!}{8}$
$The number of arrangements in which C, E, H, I, S appear in$
${that order}^{11} C_{5} \frac{6!}{2! 2! 2!} = \frac{11!}{8 \cdot 5!}$
$∴ Required Probability = \frac{\frac{11!}{8 \cdot 5!}}{\frac{11!}{8}} = \frac{1}{120}$

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