The number of bijective function $f\left(1,3,5,7,\cdots ,99\right)\to \left(2,4,6,8,\cdots ,100\right)$ if $f\left(3\right)>f\left(5\right)>f\left(7\right)\cdots >f\left(99\right)$ is

Given, $f\left(1,3,5,7,\cdots ,99\right)\to \left(2,4,6,8,\cdots ,100\right)$

The number of elements in domain and codomain is $99$.

Now, let us assume $f\left(1\right)=100$

Now as per diagram we have only $1$ way for arranging $f\left(3\right)>f\left(5\right)>f\left(7\right)\cdots >f\left(99\right)$,

Similarly if we choose $f\left(1\right)=98$ then again we have only $1$ way for arranging  $f\left(3\right)>f\left(5\right)>f\left(7\right)\cdots >f\left(99\right)$,

So we can see the arrangement $f\left(3\right)>f\left(5\right)>f\left(7\right)\cdots >f\left(99\right)$ depends upon $f\left(1\right)$,

So number of ways for choosing $f\left(1\right)$ is ${}^{50}C_1$