Let X be a set with exactly 5 elements and Y | vedclass

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(A) The number of one-one functions $\alpha$ from a set $X$ $(|X|=5)$ to a set $Y$ $(|Y|=7)$ is given by $P(7, 5) = \frac{7!}{2!} = 7 	imes 6 	imes

Vedclass · Accepted Answer

$119$

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(A) The number of one-one functions $\alpha$ from a set $X$ $(|X|=5)$ to a set $Y$ $(|Y|=7)$ is given by $P(7, 5) = \frac{7!}{2!} = 7 	imes 6 	imes 5 	imes 4 	imes 3 = 2520$.Alternatively,$\alpha = {}^{7}C_{5} 	imes 5! = 21 	imes 120 = 2520$.For onto functions $\beta$ from $Y$ $(|Y|=7)$ to $X$ $(|X|=5)$,we use the formula for the number of onto functions: $m! 	imes S(n, m)$,where $S(n, m)$ is the Stirling number of the second kind.$\beta = 5! 	imes S(7, 5) = 120 	imes \frac{1}{2!} \sum_{k=0}^{5} (-1)^{5-k} {}^{5}C_{k} k^{7} = 120 	imes 140 = 16800$.Now,calculate $\frac{1}{5!}(\beta - \alpha) = \frac{16800 - 2520}{120} = \frac{14280}{120} = 119$.

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.

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