5 men and 4 women are seated in a row. If the | vedclass

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(A) Total number of people = $5 + 4 = 9$. Total arrangements = $9!$. To find $\alpha$ (arrangements where one particular man and one particular woman

Vedclass · Accepted Answer

$2: 7$

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(A) Total number of people = $5 + 4 = 9$. Total arrangements = $9!$. To find $\alpha$ (arrangements where one particular man and one particular woman are together),treat them as a single unit. Now we have $8$ units to arrange,which can be done in $8!$ ways. Within the unit,the man and woman can be arranged in $2!$ ways. So,$\alpha = 8! 	imes 2!$. To find $\beta$ (arrangements where they are not together),subtract $\alpha$ from the total arrangements: $\beta = 9! - (8! 	imes 2!) = 9 	imes 8! - 2 	imes 8! = 7 	imes 8!$. Now,$\alpha: \beta = (8! 	imes 2) : (7 	imes 8!) = 2 : 7$.

$5$ men and $4$ women are seated in a row. If the number of arrangements in which one particular man and one particular woman are together is $\alpha$ and the number of arrangements in which those two are not together is $\beta$,then $\alpha: \beta=$

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