The number of ways in which five identical ba | vedclass

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(C) Since the $5$ balls are identical and the $10$ boxes are identical,distributing the balls such that no box contains more than one ball is equivale

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$\frac{10!}{(5!)^2}$

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(C) Since the $5$ balls are identical and the $10$ boxes are identical,distributing the balls such that no box contains more than one ball is equivalent to selecting $5$ boxes out of $10$ to place the balls in.Because the boxes are identical,the arrangement does not depend on the specific labels of the boxes,but rather on the selection of the positions.However,in standard combinatorial problems of this type where boxes are considered distinct positions,the number of ways is given by the combination formula $C(n, r) = \binom{n}{r}$.If the boxes are truly identical (indistinguishable),there is only $1$ way to place the balls. Given the options provided,the question implies the boxes are distinct positions,leading to the calculation $\binom{10}{5} = \frac{10!}{5!5!}$.

The number of ways in which five identical balls can be distributed among ten identical boxes such that no box contains more than one ball is:

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