Given, f(1,3,5,7,⋯,99)→(2,4,6,8,⋯,100)
The number of elements in domain and codomain is 99.
Now, let us assume f(1)=100

Now as per diagram we have only 1 way for arranging f(3)>f(5)>f(7)⋯>f(99),
Similarly if we choose f(1)=98 then again we have only 1 way for arranging f(3)>f(5)>f(7)⋯>f(99),
So we can see the arrangement f(3)>f(5)>f(7)⋯>f(99) depends upon f(1),
So number of ways for choosing f(1) is C150