QUESTION

# If $n\left( A \right)+n\left( B \right)=m$ , then the number of possible bijections from $A$ to $B$ .(a) $\left( \dfrac{m}{2} \right)!$(b) ${{m}^{2}}$(c) $m!$(d) $2m$

Hint: For solving this question we should know the concept of bijective functions and formula for the total number of permutations of a certain number of different objects. Then, we will solve for the correct answer.

Given:
We have two sets $A$ having $n\left( A \right)$ elements and $B$ having $n\left( B \right)$ elements. Where, $n\left( A \right)+n\left( B \right)=m$ .
Now, we have to find the total number of bijections from $A$ to $B$ . First, we see what the meaning of bijective function is. Before we see the definition of the bijective function we should know the definition of one-one and onto functions.
One-one Functions:
A function $f:A\to B$ is one-one if for any $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)\Rightarrow {{x}_{1}}={{x}_{2}}$ , i.e. the image of a distinct element of $A$ under $f$ mapping (function) are distinct. In simple words for any input value in the function there is a unique output value then, such function will be called one-one function.
Onto Functions:
A function $f:A\to B$ is onto if the range of the function $f=B$ , i.e. $f\left( A \right)=B$ , i.e. every element of $B$ is the image of some element of $A$ . In simple words the range of the function $f$ should be equal to the set of co-domain $B$ then, the function will be called onto function.
Now, functions which are both on-one and onto then, such functions are called as bijective functions. And in bijective functions, the number of elements in the domain and co-domain should be equal.
Now, it $f:A\to B$ is a bijective function where, $A$ is the domain and $B$ is the co-domain of the function then, number of elements in set $A$ i.e. $n\left( A \right)$ should be equal to the number of elements in the set $B$ i.e. $n\left( B \right)$ . Then,
\begin{align} & n\left( A \right)+n\left( B \right)=m \\ & \Rightarrow n\left( A \right)=n\left( B \right)=\dfrac{m}{2} \\ \end{align}
Now, let $A=\left\{ {{a}_{1}},{{a}_{2}},{{a}_{3}},..............,{{a}_{{}^{m}/{}_{2}}} \right\}$ and $B=\left\{ {{b}_{1}},{{b}_{2}},{{b}_{3}},..............,{{b}_{{}^{m}/{}_{2}}} \right\}$ . And if $f\left( A \right)=B$ is a bijective function then, the total number of functions can be simply calculated as we find the total number of permutations of $\dfrac{m}{2}$ distinct persons on $\dfrac{m}{2}$ distinct chairs. Then, for first-person there are $\dfrac{m}{2}$ chairs and for the second person there are $\left( \dfrac{m}{2}-1 \right)$ chairs and for the last person, there will be one chair left.
Now, the total number of permutations will be $=\dfrac{m}{2}\times \left( \dfrac{m}{2}-1 \right)\times \left( \dfrac{m}{2}-2 \right)\times ....................\times 2\times 1=\left( \dfrac{m}{2} \right)!$ .
Thus, the total number of bijective functions that are possible from $A$ to $B$ will be $\left( \dfrac{m}{2} \right)!$ .
Hence, (a) is the correct option.

Note: Here, the student should apply the condition of a function to be bijective correctly and then use the concept of permutations with the help of practical examples like we used the example of person and chairs. Then, solve for the correct answer.