Question

The number of bijective functions $f :\{1,3,5$, $7, \ldots \ldots 99\} \rightarrow\{2,4,6,8, \ldots \ldots , 100\}$, such that $f (3) \geq f (9) \geq f (15) \geq f (21) \geq \ldots f (99), $ is _____

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

Answer 1

The Correct Option is B

$f:\{1,3,5,7,\dots ..99\}\to \{2,4,6,8,\dots .,100\}$
$f(3)\ge f(9)\ge f(15)\ge \dots \dots .f(99)\dots ..\text{(1)}$
$3,9,15,\dots ..99\Rightarrow 17$ numbers
for condition one we have ${}^{50}{C}_{17}\times 1$ way rest $33$ elements $33!$
$={}^{50}{C}_{17}\times 33!$
$={}^{50}{C}_{33}\times 33!$
$={}^{50}{P}_{33}.$