Question

The number of bijection from the set $A$ to itself when $A$ contains $106$ elements is

• A. $106$
• B. ${106}^2$
• C. $106!$
• D. $2^{106}$

Answer 1

Answer: (C)

$106!$

Given $n\left(A\right)=106$
The first element of A can be mapped to any of the $106$ elements of the range ; i.e $A$

Similarly second element can be mapped to remaining $105$ elements of A
extending, we get
No of bijections from $A$ to $A$ is $106\times 105\times 104\times 103\times ......\times 3\times 2\times 1$
$=106!$