A) \[1-\frac{(n-1)!}{{{n}^{n-1}}}\]
B) \[\frac{(n-1)!}{{{n}^{n}}}\]
C) \[1-\frac{(n-1)!}{{{n}^{2}}}\]
D) None of these
Correct Answer: A
Solution :
[a] The first object can be given to any of the n persons. But the second, third and other objects, too, can go to any of the n persons. Therefore the total number of ways of distributing the n objects randomly among n persons is \[{{n}^{n}}\]. There are \[^{n}{{P}_{n}}=n!\] ways in which each person gets exactly one object, so the probability of this happening is \[\frac{n!}{{{n}^{n}}}=\frac{(n-1)!}{{{n}^{n-1}}}\]. Hence the probability that at least one person does not get any object is \[1-\frac{(n-1)!}{{{n}^{n-1}}}.\]
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