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

पर्याय

• \[\frac{1}{n^n}\]

   

• \[\frac{1}{n!}\]

   

•   \[\frac{\left( n - 1 \right)!}{n^{n - 1}}\]

   

•   None of these                             

   

Answer 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

∴ 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 3rd element in set A = n − 2

.           .           .             .             .             .             .            .

.           .           .             .             .             .             .            .

Number of ways to map nth element in set A = 1

Total number of one to one mappings from set A to itself = \[n \times \left( n - 1 \right) \times \left( n - 2 \right) \times . . . \times 1 = n!\]

∴ Required probability = \[= \frac{\text{ Total number of one to one mappings from set A to it self } }{\text{ Total number of mappings from set A to it self} } = \frac{n!}{n^n} = \frac{\left( n - 1 \right)!}{n^{n - 1}}\]