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Question: The number of surjective functions from \[A\] to \[B\] where \[A=\\{1,2,3,4\\}\] and \[B=\\{a,b\\}\]...

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

Explanation

Solution

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, ∑r=1n(−1)n−rnCrrm\sum\limits_{r=1}^{n}{{{(-1)}^{n-r}}^{n}{{C}_{r}}{{r}^{m}}}. Substituting the values of m=4m=4 and n=2n=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, ∑r=1n(−1)n−rnCrrm\sum\limits_{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≤n≤m1\le n\le m.
So, from the given values we can write,
The number of elements in set A is m=4m=4.
And the number of elements in set B is n=2n=2.
The condition is also fulfilled, that is, 1≤2≤41\le 2\le 4.
So, substituting the known values in the formula, we get the expression as,
∑r=12(−1)2−r2Crr4\sum\limits_{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)2−1.2C1(1)4+(−1)2−2.2C2(2)4= {{(-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)1.2!1!(2−1)!(1)+(−1)0.2!2!(2−2)!(16)= {{(-1)}^{1}}.\dfrac{2!}{1!\left( 2-1 \right)!}(1)+{{(-1)}^{0}}.\dfrac{2!}{2!\left( 2-2 \right)!}(16)
Solving further, we get,
=(−1).2!1!1!(1)+2!2!(0)!(16)= (-1).\dfrac{2!}{1!1!}(1)+\dfrac{2!}{2!\left( 0 \right)!}(16)
=(−1).2!+1.(16)= (-1).2!+1.(16)
=−2+16= -2+16
So, we get the value as,
=14= 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, n!r!(n−r)!\dfrac{n!}{r!\left( n-r \right)!}. In the above solution, we get 0!0!, it does not mean 0 rather its value is 1, so don’t get confused here and write the wrong value.