JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 9)
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 $${1 \over {5!}}(\beta - \alpha )$$ is ..................
Answer
119
Explanation
Given, X has exactly 5 elements and Y has exactly 7 elements.
$$ \therefore $$ n(X) = 5
and n(Y) = 7
Now, number of one-one functions from X to Y is
$$\alpha = {}^7{P_5} = {}^7{C_5} \times 5!$$
Number of onto functions from Y to X is $$\beta $$
1, 1, 1, 1, 3 or 1, 1, 1, 2, 2
$$ \therefore $$ $$\beta = {{7!} \over {3!4!}} \times 5! + {{7!} \over {{{(2!)}^3}3!}} \times 5!$$
$$ = ({}^7{C_3} + 3{}^7{C_3})5! = 4 \times {}^7{C_3} \times 5!$$
$$ \therefore $$ $${{\beta - \alpha } \over {5!}} = {{(4 \times {}^7{C_3} - {}^7{C_5})5!} \over {5!}}$$
$$ = 4 \times 35 - 21 = 140 - 21 = 119$$
$$ \therefore $$ n(X) = 5
and n(Y) = 7
Now, number of one-one functions from X to Y is
$$\alpha = {}^7{P_5} = {}^7{C_5} \times 5!$$
Number of onto functions from Y to X is $$\beta $$
1, 1, 1, 1, 3 or 1, 1, 1, 2, 2
$$ \therefore $$ $$\beta = {{7!} \over {3!4!}} \times 5! + {{7!} \over {{{(2!)}^3}3!}} \times 5!$$
$$ = ({}^7{C_3} + 3{}^7{C_3})5! = 4 \times {}^7{C_3} \times 5!$$
$$ \therefore $$ $${{\beta - \alpha } \over {5!}} = {{(4 \times {}^7{C_3} - {}^7{C_5})5!} \over {5!}}$$
$$ = 4 \times 35 - 21 = 140 - 21 = 119$$
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