Q51 / 150 · MHT CET 2012 · Mathematics

All letters of the word 'CEASE' are arranged randomly in a row, then the probability that two $\mathrm{E}$ are found together is

A
$\frac{7}{5}$
B
$\frac{3}{5}$
C
$\frac{2}{5}$
Correct answer
D
$\frac{1}{5}$

Solution

Sample space arrangement for the word 'CEASE' $=5$ ! $$ \text { Here, } \mathrm{E} \rightarrow 2, \mathrm{C} \rightarrow 1, \mathrm{~A} \rightarrow 1, \mathrm{~S} \rightarrow 1 $$ Now, consider $2 \mathrm{E}$ as one character, so that arranging for four letters $=4 !$ ways $\therefore$ Required probability $=\frac{4 !}{5 !}=\frac{1}{5}$

Hence, the correct answer is option: 'D'.

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