JEE MAIN - Mathematics (2022 - 25th July Evening Shift - No. 3)
The number of bijective functions $$f:\{1,3,5,7, \ldots, 99\} \rightarrow\{2,4,6,8, \ldots .100\}$$, such that $$f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots . . f(99)$$, is ____________.
$${ }^{50} P_{17}$$
$${ }^{50} P_{33}$$
$$33 ! \times 17$$!
$$\frac{50!}{2}$$
Explanation
As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction $$f(3) > f(9) > f(15)\,.......\, > f(99)$$
So number of ways $$ = {}^{50}{C_{17}}\,.\,1\,.\,33!$$
$$ = {}^{50}{P_{33}}$$
Comments (0)
