Number of bijective functions f :{1,3,5, 7, ldots ldots . .99} →{2,4,6,8, ldots ldots, 100} , such

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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

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