Question

Set A has 3 elements and the set B has 4 elements. Then the number of injective mappings that can be defined from A to B is:
A. 144
B. 12
C. 24
D. 60

Answer 1

Hint: To solve this question, we should know about the injective function. Let us consider a function f mapping from A to B. The function f is known as injective function when every element in the domain A is mapped to a unique element in the range B. It means that two elements of A cannot have the same mapping in the range B. In our question, it is given that A has 3 elements in it and the set B has 4 elements. We should select three elements from A and we should map the elements of A uniquely to the three selected elements of B. We should calculate the total number of such mappings using the principles of combinations. We can write that selecting three elements from the four elements of B in ${}^{4}{{C}_{3}}$ ways. The number of ways of arranging the three elements with the three elements of A is given by the formula ${}^{3}{{P}_{3}}=3!$ ways. So, the total number of ways will be the product of the two steps.

Complete step-by-step solution:
Let us consider a function $f:A\to B$. The function f is known as injective function when every element in the domain A is mapped to a unique element in the range B. It means that two elements of A cannot have the same mapping in the range B.

In our question, we are given a domain A with the number of elements as 3 and a range B with the number of elements as 4. According to the injective function definition, we should get a unique set of 3 elements from B which can be mapped by the elements in A.
We know the concept of combinations which is selecting r elements out of the n available elements is given by the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Likewise, we can write the number of ways of arranging r elements which are selected from a set of n elements as ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.
In the first step, we are selecting 3 elements from set B which has 4 elements. We can write the number of ways from the combination formula as
Number of ways $={}^{4}{{C}_{3}}=\dfrac{4!}{3!\left( 4-3 \right)!}=\dfrac{4\times 3!}{3!}=4$
In the second step, we are mapping the three elements of A with the selected three elements of B. We can write the number of ways as
Number of ways $={}^{3}{{P}_{3}}=\dfrac{3!}{\left( 3-3 \right)!}=\dfrac{3!}{0!}=3\times 2\times 1=6$
As the two steps should happen to get the mapping, the total number of ways will be the product of both the numbers. We can write the total number of ways as
Number of ways $=4\times 6=24$
$\therefore $ There are 24 ways of mapping an injective function from A to B. The answer is option-C.

Note: An alternate way to do this problem is by combining both the steps. We are selecting three elements from the set of 4 elements of B and arranging them to the three elements of A. We can infer the direct arrangement formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ can be applied here with n = 4 and r = 3. Using them in the above formula gives
${}^{4}{{P}_{3}}=\dfrac{4!}{\left( 4-3 \right)!}=\dfrac{4!}{1!}=4\times 3\times 2\times 1=24$ which is same as the above answer.