Question

If $n\left(A\right)+n\left(B\right)=m$, then the number of possible bijections from $A$ to $B$ is

• A. $\left(\frac{m}{2}\right)!$
• B. $m^2$
• C. $m!$
• D. $2m$

Answer 1

The correct option is A $\left(\frac{m}{2}\right)!$

A bijection from $A$ to $B$ is a function which maps every element of $A$ to unique element of $B$ i.e. injective.

$⟹n\left(B\right)\ge n\left(A\right)$

Also, it ensures that every element of $B$ is an image of some element of $A$

$⟹n\left(A\right)\ge n\left(B\right)$

$\therefore n\left(A\right)=n\left(B\right)$

$⟹n\left(A\right)=\frac{m}{2}=n\left(B\right)$

Let $A=\{a_1,a_2,.....,a_{\frac{m}{2}}\}$ and

$B=\{b_1,b_2,.....,b_{\frac{m}{2}}\}$

Let $f:A\to B$ defined by $f\left(a_i\right)=b_i$ is a bijection.

Any and all images of some fixed $a_i$ appears in at least one such $f$. And each $f$ is unique for each permutation $(b_1,b_2,...b_{\frac{m}{2}})$.

Hence, the number of functions is exactly equal to the number of such permutations, which
is $\left(\frac{m}{2}\right)!$