Question

Let A be a set containing $10$ distinct elements. Then the total number of distinct functions from A to A, is?
$A)10!$
$B){10^{10}}$
$C){2^{10}}$
$D){2^{10}} - 1$

Answer 1

Solution: Hint:
First, from the given that we have” A” be a set containing $10$ distinct elements. Distinct elements mean all the elements in that function are different elements, which means no identical elements.
We also need to know about the concept of domain and co-domain as well, which is also known as the mapping in the range functions.

Complete step-by-step solution:
Since from the given question, we have A be a set containing $10$ distinct elements and the total number of distinct functions from A to A is the requirement.
Let us assume the set A with the elements like ${A_1},{A_2},...,{A_{10}}$
Then by using the function definition of the domain and co-domain we have a function from A to A which means itself: $f({A_m}) = {A_n}$ where m and n are the ranges from $1,2,...,10$
Therefore, the total number of distinct function for number $1$ is $1,2,...,10$ (it can be represented in any ten numbers) and the number $2$ is $1,2,...,10$ and proceeding like this we also get the number $10$ as $1,2,...,10$
Thus, we get for all the ten numbers we have $1,2,...,10$
Hence, we have in total $10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 = {10^{10}}$ ways.
Therefore, the option $B){10^{10}}$ is correct.

Note: The domain is defined as the set which is to input in a function. We say that input values satisfy a function. The range is defined as the actual output supposed to be obtained by entering the domain of the function.
Co-domain is defined as the values that are present in the right set that is set Y, possible values expected to come out after entering domain values.