Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If α is the number of…

Solution

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

So, $\beta =\frac{7!}{3!4!}\times 5!+\frac{7!}{{\left(2!\right)}^{3}3!}\times 5!=\left({}^7{C}_3+3.{ }^7{C}_3\right)5!=4{\times }^7{C}_3\times 5!$
$\frac{\beta -\alpha }{5!}=4\times {}^7{C}_3-{}^7{C}_5=4\times 35-21=119$

Asked in: JEE Advanced 2018 (Paper 2)

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