If 10 different balls are to be placed in 4 d | vedclass

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(C) Total ways to place $10$ distinct balls in $4$ distinct boxes is $4^{10}$.To find the number of ways such that two boxes contain exactly $2$ and $

Vedclass · Accepted Answer

$\frac{945}{2^{10}}$

Vedclass · Answer

(C) Total ways to place $10$ distinct balls in $4$ distinct boxes is $4^{10}$.To find the number of ways such that two boxes contain exactly $2$ and $3$ balls:$1$. Select $2$ boxes out of $4$ to contain $2$ and $3$ balls respectively: $P(4, 2) = 4 	imes 3 = 12$ ways.$2$. Select $2$ balls for the first box and $3$ balls for the second box from $10$ balls: $\binom{10}{2} 	imes \binom{8}{3} = 45 	imes 56 = 2520$ ways.$3$. The remaining $5$ balls can be placed in the remaining $2$ boxes in $2^5 = 32$ ways.Total favorable ways $= 12 	imes 2520 	imes 32 = 967680$.Probability $= \frac{967680}{4^{10}} = \frac{967680}{1048576} = \frac{945}{1024} = \frac{945}{2^{10}}$.

If $10$ different balls are to be placed in $4$ distinct boxes at random,then the probability that two of these boxes contain exactly $2$ and $3$ balls is

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