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1.Relation and Function
hard
The number of bijective functions $f :\{1,3,5, 7, \ldots \ldots . .99\} \rightarrow\{2,4,6,8, \ldots \ldots, 100\}$, such that $f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots \ldots f(99), \quad$ is
A
${ }^{50} P _{17}$
B
${ }^{50} P _{33}$
C
$33 ! \times 17 !$
D
$\frac{50 !}{2}$
(JEE MAIN-2022)
Solution
As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction
$f(3)>f(9)>f(15) \ldots . .>f(99)$
So number of ways $={ }^{50} C_{17} \cdot 1 \cdot 33$ !
$={ }^{50} P_{33}$
Standard 12
Mathematics
