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(C) The 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 togeth

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(C) The 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 $4$ units can be arranged in $4!$ ways.Within the unit $(AE)$,$A$ and $E$ can be arranged in $2!$ ways.So,the total favourable ways $= 2! 	imes 4! = 2 	imes 24 = 48$.The required probability $= \frac{	ext{Favourable ways}}{	ext{Total ways}} = \frac{48}{120} = \frac{2}{5}$.

$5$ 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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