Question

A mapping is selected at random from set $A=\{1,2,.10\}$ into itself. The probability that mapping selected is an injective, is?

• A. $\frac{10}{{10}^9}$
• B. $\frac{9!}{{10}^9}$
• C. $\frac{9}{10!}$
• D. None of these

Answer 1

The correct option is B $\frac{9!}{{10}^9}$

Given set $A=\{1,2,....,10\}$ which is mapped to it self.

For the mapping to be injective each and every

element of A is to be mapped such that no 2

elements are mapped to the same element

The first element of A can be mapped to any of the

10 elements. But for injective mapping the 2${}^{nd}$

element should not be mapped to this element

making the number of possibilities as 9. similar

way for other.

So, Number of possible invective mapping

$=10\times 9\times 8\times ...\times 2\times 1$

$=10!$

Total number of mappings $=10\times 10\times ...\times 10\left(10times\right)$

${10}^{10}$

Probability that mapping is injective $=\frac{10!}{{10}^{10}}$

$=\frac{9!}{{10}^9}$