Question

The number of surjective functions from $$A$$ to $$B$$ where $$A=\{1,2,3,4\}$$ and $$B=\{a,b\}$$ is
A. 14
B. 12
C. 5
D. 15

Answer 1

Hint: In the given question, we are given two sets namely, A and B and using these given sets we have to find the number of surjective functions. To calculate the number of surjective function, we will be using the formula, $$\sum _{r=1}^n{{(-1)}^{n-r}}^n{C}_r{r}^m$$. Substituting the values of $$m=4$$ and $$n=2$$ in the given expression, we will get the value of the number of surjective functions.

Complete step by step solution:
According to the given question, we are given two sets namely, A and B and using these sets given to us we have to find the number of surjective functions.
Surjective function can be defined as a function f from a set X to a set Y, if every element in Y (codomain) has at least one element in the X. Uniqueness is not a necessity in this case.
The formula for finding the number of surjective function is, $$\sum _{r=1}^n{{(-1)}^{n-r}}^n{C}_r{r}^m$$
Where m and n are the number of the elements of the sets X and Y respectively such that $$1\le n\le m$$.
So, from the given values we can write,
The number of elements in set A is $$m=4$$.
And the number of elements in set B is $$n=2$$.
The condition is also fulfilled, that is, $$1\le 2\le 4$$.
So, substituting the known values in the formula, we get the expression as,
$$\sum _{r=1}^2{{(-1)}^{2-r}}^2{C}_r{r}^4$$
Expanding the above expression further and solving it, we get the value as,
$$={(-1)}^{2-1}{.}^2{C}_1{(1)}^4+{(-1)}^{2-2}{.}^2{C}_2{(2)}^4$$
Using the formula of the combination to expand the expression, we have,
$$={(-1)}^1.{\displaystyle \frac{2!}{1!(2-1)!}}(1)+{(-1)}^0.{\displaystyle \frac{2!}{2!(2-2)!}}(16)$$
Solving further, we get,
$$=(-1).{\displaystyle \frac{2!}{1!1!}}(1)+{\displaystyle \frac{2!}{2!\left(0\right)!}}(16)$$
$$=(-1).2!+1.(16)$$
$$=-2+16$$
So, we get the value as,
$$=14$$

So, the correct answer is “Option A”.

Note: The formula to find the number of surjective functions between two sets has many components to it and so has a very good chance of writing the expression incorrectly. Also, the combinations formula applied in the above solution should also be carefully done, which is, $${\displaystyle \frac{n!}{r!(n-r)!}}$$. In the above solution, we get $$0!$$, it does not mean 0 rather its value is 1, so don’t get confused here and write the wrong value.