Question

The number of one-one functions that can be defined from A = {4, 8, 12, 16} to B is 5040, then n(B)=
a). 7
b). 8
c). 9
d). 10

Answer 1

Hint: For solving this problem, we first find the number of elements given for set A. The number of one-one functions from A to B is 5040. Let the number of elements in B be n. Now by applying the formula for one-one functions that is ${}^{n}{{P}_{m}}$, we can obtain the number of elements in B.

Complete step-by-step solution -
A function as is said to be a one-one function or an injection, if different elements of A have different images in B. If A and B are finite sets having m and n elements respectively such that m is less than equal to n, then to define a one-one function from A to B, we have to relate m elements in A to n distinct elements in B. Thus, number of one-one function: ${}^{n}{{P}_{m}}$.
According to our problem we are given that the number of functions that can be defined from A to B is 5040. The number of elements in A are 4. We are required to find the number of elements in B. By using the above formula and putting m = 4 and n same, we get
$ \Rightarrow {}^{n}{{P}_{m}}=5040 \\ $
$ \Rightarrow {}^{n}{{P}_{4}}=5040 \\ $
From the definition of permutation, ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
On expanding further, we get
$\Rightarrow {}^{n}{{P}_{4}}=\dfrac{n!}{\left( n-4 \right)!} \\ $
$\Rightarrow \dfrac{n!}{\left( n-4 \right)!}=5040 \\ $
As, we know $\ {}^{10}{{P}_{4}}=5040 \\ $
$\Rightarrow \dfrac{n!}{\left( n-4 \right)!}=\dfrac{10!}{\left( 10-4 \right)!} \\ $
$ \therefore n=10 \\ $
Therefore, there are 10 elements present in B.
Hence, option (d) is correct.

Note: Another possible method to solve this problem can be explained as: Let the first element 4 in A be mapped on n elements of B. Similarly, the second element 8 is mapped on (n -1) elements. Hence, we can write for four elements mapping as: $n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)=5040$. Putting n as 10, we get the same result.