Question
If $$n$$ objects are distributed at random among $$n$$ persons, the probability that at least one of them will not get anything is :
A.
$$1 - \frac{{\left( {n - 1} \right)!}}{{{n^{n - 1}}}}$$
B.
$$\frac{{\left( {n - 1} \right)!}}{{{n^n}}}$$
C.
$$1 - \frac{{\left( {n - 1} \right)!}}{{{n^n}}}$$
D.
none of these
Answer :
$$1 - \frac{{\left( {n - 1} \right)!}}{{{n^{n - 1}}}}$$
Solution :
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 $$\frac{{n!}}{{{n^n}}} = \frac{{\left( {n - 1} \right)!}}{{{n^{n - 1}}}}.$$
Hence the probability that at least one person does not get any object is $$1 - \frac{{\left( {n - 1} \right)!}}{{{n^{n - 1}}}}.$$