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

Answer: (B)

${ }^{50} P _{33}$

Answer 2

f:{1,3,5,7,…..99}→{2,4,6,8,….,100}
f(3)≥f(9)≥f(15)≥…….f(99)….. (1)
3,9,15,…..99⇒17 numbers
for condition one we have 50C17​×1 way rest 33 elements 33!
=50C17​×33!
=50C33​×33!
=50P33​.