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(C) Total number of ways to arrange $5$ persons in a queue is $5! = 120$.To find the number of favourable ways where $A$ and $E$ are always together,w

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(C) Total number of ways to arrange $5$ persons in a queue is $5! = 120$.To find the number of favourable ways where $A$ and $E$ are always together,we treat $(AE)$ as a single unit.Now,we have $4$ units to arrange: $(AE), B, C, D$. These can be arranged in $4!$ ways.Within the unit $(AE)$,$A$ and $E$ can be arranged in $2! = 2$ ways.So,the total number of favourable ways is $2 	imes 4! = 2 	imes 24 = 48$.The required probability is $\frac{	ext{Favourable outcomes}}{	ext{Total outcomes}} = \frac{2 	imes 4!}{5!} = \frac{2 	imes 24}{120} = \frac{48}{120} = \frac{2}{5}$.

$5$ persons $A, B, C, D$ and $E$ are in a queue at a shop. The probability that $A$ and $E$ are always together is:

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