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All letters of the word 'CEASE' are arranged randomly in a row, then the probability that two $\mathrm{E}$ are found together is
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Verified Answer
The correct answer is:
$\frac{1}{5}$
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 $=\frac{4 !}{5 !}=\frac{1}{5}$
$$
\text { Here, } \mathrm{E} \rightarrow 2, \mathrm{C} \rightarrow 1, \mathrm{~A} \rightarrow 1, \mathrm{~S} \rightarrow 1
$$
Now, consider $2 \mathrm{E}$ as one character, so that arranging for four letters $=4 !$ ways
$\therefore$ Required probability $=\frac{4 !}{5 !}=\frac{1}{5}$
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