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# 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. $\dfrac{1}{{{n^n}}}$ b. $\dfrac{1}{{n!}}$c. $\dfrac{{(n - 1)!}}{{{n^{n - 1}}}}$d. None of these.  Hint: Here in this question you need to know two concepts one is probability and other is one to one mapping.
Probability: -
Probability=Favourable outcomes/Total number of outcomes.
Probability can never be greater than one.
One to one mapping: -
One to one or (1-1) function is a relation that preserves the “uniqueness”. Every unique member of the function’s domain is mapped to the unique member of the function’s range. This mapping is sometimes also called as injective mapping.

Complete step-by-step answer: Here given set is A=1, 2, 3, ...., n
Number of ways first element of set A can be mapped = n
Number of ways second element of set A can be mapped = n
Number of ways the third element of set A can be mapped = n and so on.
Total number of mapping from set A to itself=$n \times n \times ...... \times n = {n^n}$ (it is being multiplied to n times)
For one to one mapping: -
Number of ways to map first element in set A=n
Number of ways to map second element in set A=n-1
Number of ways to map third element in set A=n-2
Number of ways to map nth element in set A=1
Total number of one to one mapping from set A to itself=$n \times (n - 1) \times (n - 2)..... \times 1 = n!$
$\therefore$ Required probability=Total number of one to one mapping from set A to itself/Total number of mapping from set A to itself
$\Rightarrow \dfrac{{n!}}{{{n^n}}} = \dfrac{{n(n - 1)!}}{{{n^n}}}$ (Opening factorial using formula$n! = n(n - 1)(n - 2)(n - 3) \times ....... \times 3 \times 2 \times 1$ )
$\therefore \dfrac{{(n - 1)!}}{{{n^{n - 1}}}}$ (Using identity $\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}$ )
Hence option C is correct.

Note: Most of the times students got confused in one to one mapping so here is some easy way to understand it using this diagram. View Notes
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