Question

Let $X$ be a set with exactly $5$ elements and $Y$ be a set with exactly $7$ elements. If $\alpha$ is the number of one-one functions from $X$ to $Y$ and $\beta$ is the number of onto functions from $Y$ to $X,$ then the value of $\frac{1}{5!}(\beta -\alpha )$ is

Answer 1

$n\left(X\right)=5$ and $n\left(Y\right)=7$
$\alpha \to \text{number of one-one function from}X\text{to}Y\alpha ={}^7{C}_5\times 5!$
$\beta \to \text{number of onto function from}Y\text{to}X$

For onto function, Co-domain and range of the function should be equal therefore 2 cases are possible:
Case $1:3$ elements of $Y$ are mapped to any one element of $X$ and remaining 4 elements of $X$ and $Y$ are mapped with each other.
Case $2:4$ elements of $Y$ are mapped to any two elements of $X$ and remaining 3 elements of $X$ and $Y$ are mapped with each other.
$\therefore \beta =[{}^7{C}_3+\frac{{}^7{C}_2.{}^5{C}_2}{2!}]5!={}^7{C}_3.4.5!$
Hence, $\frac{\beta -\alpha }{5!}=\frac{({}^7{C}_3.4-{}^7{C}_5)5!}{5!}=119$


$\text{Alternate Solution}$

Number of one-one functions from $\left(X\right(n\left(X\right)=n=5\left)\right)$ to $\left(Y\right(n\left(Y\right)=m=7\left)\right)$
$\alpha ={}^m{C}_n\times n!(\because m\ge n)$
$={}^7{C}_5\times 5!=2520$

Number of onto functions from $\left(Y\right(n\left(Y\right)=m=7\left)\right)$ to $\left(X\right(n\left(X\right)=n=5\left)\right)$
$\because m>n$
$\therefore \beta =n^m-{(}^n{C}_1(n-1{)}^m-{}^n{C}_2(n-2{)}^m+\dots )$
$=5^7-{(}^5{C}_1\cdot 4^7-{}^5{C}_2\cdot 3^7+{}^5{C}_3\cdot 2^7-{}^5{C}_4\cdot 1^7)$
$=78125-(81920-21870+1280-5)$
$=16800$

Now, $\frac{\beta -\alpha }{5!}=\frac{16800-2520}{120}$
$=\frac{14280}{120}=119$