If the letter of the word $A S H O K A$ are written down at randomly, then the chance that all A's are consecutive is

\frac{1}{3}

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\frac{1}{4}

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\frac{2}{3}

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\frac{1}{2}

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Solution

The correct option is C $\frac{1}{3}$
Number of ways of writing $A S H O K A$ , where the all A's are consecutive $= 5!$
And total number of ways of writing $A S H O K A = \frac{6!}{2!}$
( $∵$ Here, number of times $A$ repeated is $2$ )
$∴$ Required chance $= \frac{5!}{(\frac{6!}{2!})} = \frac{1}{3}$

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