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Probability

If n distin...

Question

If $n$ distinct balls are placed in $n$ cells, the probability that each cell will be occupied is

\frac{1}{n^{n - 1}}

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\frac{1}{n^{n}}

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\frac{n!}{n^{n}}

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1 - \frac{p!}{n^{n - 1}}

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Solution

The correct option is C $\frac{n!}{n^{n}}$
The ways of placing $n$ balls in $n$ cells such that each get one $= n!$
The ways of placing $n$ balls in $n$ cells $= n^{n}$
The required probability $= \frac{n!}{n^{n}}$

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