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(D) The total number of ways to place $8$ balls in $8$ bags is $8!$. We want to find the number of ways such that exactly $5$ balls are in their respe

Vedclass · Accepted Answer

$\frac{1}{360}$

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(D) The total number of ways to place $8$ balls in $8$ bags is $8!$. We want to find the number of ways such that exactly $5$ balls are in their respective coloured bags. First,we choose $5$ balls out of $8$ to be in their correct bags,which can be done in $\binom{8}{5}$ ways. The remaining $3$ balls must be placed in the remaining $3$ bags such that none of them is in its respective coloured bag. This is a derangement of $3$ items,denoted by $D_3$. $D_3 = 3! 	imes (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}) = 6 	imes (1 - 1 + 0.5 - 0.1667) = 2$. So,the number of favorable outcomes is $\binom{8}{5} 	imes D_3 = \binom{8}{3} 	imes 2 = 56 	imes 2 = 112$. The probability is $\frac{112}{8!} = \frac{112}{40320} = \frac{1}{360}$.

There are $8$ different coloured balls and $8$ bags having the same colours as that of the balls. If one ball is placed at random in each one of the bags,then the probability that $5$ of the balls are placed in the respective coloured bags,is

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