Question

One mapping is selected at random from all the mapping 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. $\frac{(n-1)!}{n^{n-1}}$
• D. None of these

Answer 1

The correct option is C

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

Number of ways to map 1st element in set A = n

Number of ways to map 2nd element in set A = n and so on

$\therefore$ Total number of mapping from set A to itself $=n\times n\times .....\times n$ (n times) = $n^n$

For one to one mapping,

Number of ways to map 1st element in set A = n

Number of ways to map 2nd element in set A = n - 1

Number of ways to map 3nd element in set A = n - 2

Number of ways to map $n^{th}$ element in set A = 1

Total number of one to one mapping from set A to itself $=n\times (n-1)\times (n-2)\times ....\times 1=n!$

$\therefore$ Required probability

Total number of one to one

\(= \frac{\text{mappings from set A to itself}}{\text{Total number of mapping from }\)

set A to itself

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