The number of bijective functions $$f(\{1, 3, 5, 7, \ldots, 99\}) \to \{2, 4, 6, 8, \ldots, 100\}$$ if $$f(3) > f(5) > f(7) \ldots > f(99)$$ is

We wish to count bijective functions $$f:\{1,3,5,7,\ldots,99\}\to\{2,4,6,8,\ldots,100\}$$ satisfying the strict inequality $$f(3)>f(5)>f(7)>\cdots>f(99)\,.$$

Both the domain and the codomain consist of 50 elements, namely $$\{1,3,5,\ldots,99\}$$ and $$\{2,4,6,\ldots,100\}$$ respectively. Since $$f$$ is bijective, every one of the 50 codomain values must be used exactly once.

The requirement that the values assigned to the 49 inputs $$\{3,5,7,\ldots,99\}$$ form a strictly decreasing sequence implies that once we choose which 49 codomain values fill those inputs, there is exactly one way to order them so that $$f(3)>f(5)>\cdots>f(99)\,. $$ The remaining input, 1, can be mapped to any of the 50 codomain values; choosing its image then determines exactly which 49 values remain for the decreasing chain.

Therefore, the total number of such bijections is simply the number of choices for $$f(1)$$, namely:

The answer is Option A.