Question

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

• A. $33!\times 17!$
• B. ${}^{50}{P}_{33}$
• C. ${}^{50}{P}_{17}$
• D. $\frac{50!}{2}$

Answer 1

The correct option is B ${}^{50}{P}_{33}$

As fuunction is one-one and onto, out of $50$ elements of domain set $17$ elements are following restriction
$f\left(3\right)\ge \left(f\right(9)\ge (f\left(15\right))...\ge (f\left(99\right))$
So number of ways $={}^{50}{C}_{17}33!$
$={}^{50}{P}_{33}$