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

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(B) 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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(B) The number of one-one functions $\alpha$ from a set $X$ $(|X|=5)$ to a set $Y$ $(|Y|=7)$ is given by $P(7, 5) = \frac{7!}{2!} = 7 	imes 6 	imes 5 	imes 4 	imes 3 = 2520$.The number of onto functions $\beta$ from a set $Y$ $(|Y|=7)$ to a set $X$ $(|X|=5)$ is given by the formula $5! 	imes S(7, 5)$,where $S(7, 5)$ is the Stirling number of the second kind.$S(7, 5) = \frac{1}{5!} \sum_{k=0}^{5} (-1)^k \binom{5}{k} (5-k)^7 = \frac{1}{120} [1 	imes 5^7 - 5 	imes 4^7 + 10 	imes 3^7 - 10 	imes 2^7 + 5 	imes 1^7] = 140$.Thus,$\beta = 120 	imes 140 = 16800$.We need to 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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