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Question

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 ____________.

${ }^{50} P_{17}$
${ }^{50} P_{33}$
$33 ! \times 17$!
$\frac{50!}{2}$

Solution

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