Question

From the set {a, b, c, d} to set {1, 2, 3, 4} , what’s the total number of one-one functions that can be found?

Answer 1

Hint:Use the cardinality of two sets to find the number of one-one functions between them.
\[\text{Number of one-one functions = }{}^{n}{{P}_{m}}\text{ if n}\ge \text{m}\]…..(1)
Number of one-one functions = 0 if n < m…..(2)


Complete step-by-step solution:
The total number of elements in the set is called the cardinality of the set.
Let us assume given sets as A and B, that is,
A = {a, b, c, d} and B = {1, 2, 3, 4}
The cardinality of a set is denoted by “|set|”
Here cardinality of A = |A| = 4.
Cardinality of B = |B| = 4.
If there are two non-empty sets with cardinality m and n, then the number of one-one functions between them is given by:
\[\text{Number of one-one functions = }{}^{n}{{P}_{m}}\text{ if n}\ge \text{m}\] …..(1)
Number of one-one functions = 0 if n < m…..(2)
By the above formula, in our case the value of m is 4 and the value of n is 4.
We can see that m = n.
So we need to use equation (1):
\[\text{Number of one-one functions = }{}^{n}{{P}_{m}}\text{ if n}\ge \text{m}\].
So, the number of one-one functions = \[{}^{4}{{P}_{4}}\]
By using the formula,
\[{}^{n}{{P}_{m}}=\dfrac{n!}{\left( n-m \right)!}\]
By substituting factorial of 0 as 1,we get:
\[\text{The number of one-one functions }={}^{4}{{P}_{4}}=\dfrac{4!}{(4-4)!}=4!\] [$\because$ 0!=1]

The number of one-one functions = (4)(3)(2)(1) = 24.
\[\therefore \]The total number of one-one functions from {a, b, c, d} to {1, 2, 3, 4} is 24.



Note: Here the values of m, n are same but in case they are different then the direction of checking matters. If m > n, then the number of one-one from first set to the second becomes 0. So take care of the direction of checking.