Question

One mapping is selected at random from all mappings of the set A = {1, 2, 3, ..., n} into itself. The probability that the mapping selected is one to one is

(a) $\frac{1}{n^n}$ (b) $\frac{1}{n!}$ (c) $\frac{\left(n-1\right)!}{n^{n-1}}$ (d) None of these

Answer 1

For a set with n element, say A
Total number of mapping from a set having n elements is nⁿ.
For one to one mapping the first element is A can have any of the n images A.
Second element in A can have any of remaining (n − 1) images.
Continuing like this
n^th element will have 1 option left
∴ Total number of one to one mapping is n!
$$\begin{gathered} \mathrm{Hence}\mathrm{required}\mathrm{probability}\mathrm{is}\frac{n!}{n^n} \\ =\frac{n\left(n-1\right)!}{n{n}^{n-1}} \\ =\frac{\left(n-1\right)!}{n^{n-1}} \end{gathered}$$
​Hence, the correct answer is option C.