Question

A mapping is selected at random from the set of all the mappings of the set $A=\{1,2,..,n\}$ into itself. The probability that the mapping selected is an injection is

• A. $\frac{(n-1)!}{n^n}$
• B. $\frac{(n-1)!}{n^{n-1}}$
• C. $\frac{n-1}{n^{n-1}}$
• D. $\frac{n}{n^{n-1}}$

Answer 1

The correct option is B $\frac{(n-1)!}{n^{n-1}}$

Total number of functions from $A$ to itself is $n^n$.
Out of these functions, $n!$ function are injective mappings (one-one).
So required probability
$=\frac{n!}{n^n}=\frac{(n-1)!}{n^{n-1}}$