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Question: One mapping is selected at random from all mappings of the set \[S = \\{ 1,2,3,....,n\\} \] into its...

One mapping is selected at random from all mappings of the set S=1,2,3,....,nS = \\{ 1,2,3,....,n\\} into itself. If the probability that mapping is one-one is 332\dfrac{3}{{32}} then the value of n is ?
A) 2
B) 3
C) 4
D) None of these

Explanation

Solution

A relation R: AB\mathbb{R}:{\text{ A}} \to {\text{B}}is defined as a function if every element of set A is mapped to one element of set B, i.e. , each object has only one image.

One-one function: When each element of set B is mapped to only one element of set A, i.e., each object in set A has a unique image in set B, then the function is called one-one function.
Let the number of elements in set A = n(A) = a
The number of elements in set B = n(B) = b
Necessary condition :  a{\text{b }} \geqslant {\text{ a}} [ for one-one function f: ABf:{\text{ A}} \to {\text{B}}] …… (1)
Total number of functions =ba= \mathop {\text{b}}\nolimits^{\text{a}} …… (2)
Number of one-one functions =C(b,a)a! = C({\text{b,a)}} \cdot {\text{a!}} (Cis combination)(\because 'C'{\text{is combination)}} …… (3)

Complete step by step solution:
The mapping of set S=1,2,3,....,nS = \\{ 1,2,3,....,n\\} into itself can be drawn as,

Where the image of every element xx in the domain, there exists an element f(x)f(x) in the range.

Step 1
Given set S=1,2,3,....,nS = \\{ 1,2,3,....,n\\}

Step 2
Function f: SSf:{\text{ S}} \to {\text{S}}is defined. (given)(\because {\text{given}})

Step 3
Number of elements in set S = n(S) = n.
Total number of functions f: SS =  nnf:{\text{ S}} \to {\text{S = }}\mathop {{\text{ }}n}\nolimits^n (from (2)) …… (4)

Step 4
Number of one-one functions f:SS = C(n,n).n!f:{\text{S}} \to {\text{S = }}C({\text{n,n)}}{\text{.n}}! (from (3))
=n!= {\text{n!}} …… (5)

Step 5
Probability (selecting mapping is one-one) = number of one - one functions f:S  total number of functions f:SS=332\dfrac{{{\text{number of one - one functions }}f:{\text{S}} \to {\text{S }}\;}}{{{\text{total number of functions }}f:{\text{S}} \to {\text{S}}}} = \dfrac{3}{{32}}
(given)(\because {\text{given}})
=n!nn=332= \dfrac{{{\text{n!}}}}{{\mathop {\text{n}}\nolimits^{\text{n}} }} = \dfrac{3}{{32}} (from (4) and (5)) \therefore n = 4 [by inspection]

Therefore, the correct option is (C), n=4.

Note:
Another classification of functions is onto functions. If for functions f:ABf:{\text{A}} \to {\text{B}}the co-domain set of B is also the range for the function, then the function is called an onto function.
The functions which are both one-one and onto are called bijective functions.