The number of bijective function f(1, 3, 5, 7, ⋯ , 99) → (2, 4, 6,...

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

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 $f\left(3\right) > f\left(5\right) > f\left(7\right) ⋯ > f\left(99\right)$,

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

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

So number of ways for choosing $f\left(1\right)$ is $C1 50$