The number of bijective functions f : {1, 3, 5, 7,........99}$\stackrel{ }{\to }${2, 4, 6, 8,......., 100}, such that $\mathrm{f}\left(3\right)\ge \mathrm{f}\left(9\right)\ge \mathrm{f}\left(15\right)\ge \mathrm{f}\left(21\right)\ge \dots ..\ge \mathrm{f}\left(99\right)$, is _____

f : {1, 3, 5, 7,........99}$\stackrel{ }{\to }${2, 4, 6, 8,......., 100},$$\begin{gathered} f:A\to B where A=\left\{1,3,5,7,,,,,99\right\},B=\left\{2,4,4,8,....100\right\} \\ n\left(A\right)=n\left(B\right)=50 \end{gathered}$$

$$\begin{gathered} \mathrm{f}\left(3\right)\ge \mathrm{f}\left(9\right)\ge \mathrm{f}\left(15\right)\ge \mathrm{f}\left(21\right)\ge \dots ..\ge \mathrm{f}\left(99\right) \\ 3,9,15,21,...99 are in A.P with a_n=99, a_1=3,d=6 \\ \Rightarrow 99=3+\left(n-1\right)6\therefore n=17 \\ \mathrm{f}\left(3\right),\mathrm{f}\left(9\right),\mathrm{f}\left(15\right),\mathrm{f}\left(21\right),\dots ..,\mathrm{f}\left(99\right) can be selected in{ }^{50}{\mathrm{C}}_{17} \mathrm{ways} \\ \mathrm{f} \mathrm{images} \mathrm{of} \mathrm{other} 33 \mathrm{elements} \mathrm{can} \mathrm{be} \mathrm{selected} \mathrm{and} \mathrm{arranged} \mathrm{in} 33! \mathrm{ways} \\ \therefore \mathrm{The} \mathrm{number} \mathrm{of} \mathrm{bijections} ={ }^{50}{\mathrm{C}}_{17}\left(33!\right){=}^{50}{\mathrm{P}}_{33} \end{gathered}$$