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 ______ .

$|X| = 5,\qquad |Y| = 7$ $\alpha = \#\{\text{one-one functions } X \to Y\}$ $\beta = \#\{\text{onto functions } Y \to X\}$ $\text{To find: } \dfrac{1}{5!}(\beta - \alpha)$ $\text{Plan: }1.\ \alpha = {}^{7}P_{5}$ $2.\ \beta = \#\{\text{onto } Y \to X\}$ $3.\ \dfrac{\beta - \alpha}{5!}$ $\alpha = \binom{7}{5}\cdot 5! = {}^{7}P_{5}$ $\alpha = 7\cdot 6\cdot 5\cdot 4\cdot 3 = 2520$ $\beta = \sum_{k=0}^{5} (-1)^{k}\binom{5}{k}(5-k)^{7}$ $= 5^{7} - \binom{5}{1}4^{7} + \binom{5}{2}3^{7} - \binom{5}{3}2^{7} + \binom{5}{4}1^{7}$ $= 78125 - 5(16384) + 10(2187) - 10(128) + 5(1)$ $\beta = 16800$ $\beta = 16800 = 5!\,S(7,5)$ $\dfrac{\beta}{5!} = S(7,5) = 140$ $\dfrac{\alpha}{5!} = \dfrac{{}^{7}P_{5}}{5!} = \dfrac{7\cdot 6}{2} = 21$ $\dfrac{\beta - \alpha}{5!} = 140 - 21 = \boxed{119}$