1. Tardigrade
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  3. Mathematics
  4. The number of bijective functions f: 1,3,5, 7, ldots ldots . .99 arrow 2,4,6,8, ldots ldots . ., 100 such that f (3) ≥ f (9) ≥ f (15) ≥ f (21) ≥ ldots . . f (99), is

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Q. 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 . . f (99), $ is _____

JEE MainJEE Main 2022Permutations and Combinations

Solution:

$ f :\{1,3,5,7, \ldots . .99\} \rightarrow\{2,4,6,8, \ldots ., 100\} $
$ f (3) \geq f (9) \geq f (15) \geq \ldots \ldots . f (99) \ldots . . \text { (1) } $
$ 3,9,15, \ldots . .99 \Rightarrow 17 $ numbers
for condition one we have ${ }^{50} C _{17} \times 1$ way rest $33$ elements $33 !$
$={ }^{50} C _{17} \times 33 ! $
$ ={ }^{50} C _{33} \times 33 ! $
$ ={ }^{50} P _{33} .$