Question

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

• A. $\frac{1}{n^n}$
• B. $\frac{1}{n!}$
• C. $\frac{(n-1)!}{n^{n-1}}$
• D. None of these

Answer 1

Answer: (C)

$\frac{(n-1)!}{n^{n-1}}$

Total no. of mappings for set A (with n elements) into itself = $n^n$
No. of one-to-one mappings for set A into itself = $n!$
Therefore, probability = $\frac{n!}{n^n}=\frac{(n-1)!}{n^{(n-1)}}$