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 \_\_\_\_\_\_\_\_\_\_\_\_.
Solutions · 1
Official solution
As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction \[f(3) > f(9) > f(15)\,.......\, > f(99)\]
So number of ways \[ = {}^{50}{C_{17}}\,.\,1\,.\,33!\]
=50P33