The number of bijective functions f: 1,3,5, 7, ldots ldots . .99 arrow 2,4,6,8, ldots ldots . ., 100 such that f (3) ≥ f (9) ≥ f (15) ≥ f (21) ≥ ldots . . f (99), is

Question

The number of bijective functions f: 1,3,5, 7, ldots ldots . .99 arrow 2,4,6,8, ldots ldots . ., 100 such that f (3) ≥ f (9) ≥ f (15) ≥ f (21) ≥ ldots . . f (99), is  (A)  50 P 17 (B)  50 P 33 (C) 33 ! × 17 ! (D) (50 text ! /2)

Tardigrade Admin · Accepted Answer

f: 1,3,5,7, ldots . .99 arrow 2,4,6,8, ldots ., 100 f (3) ≥ f (9) ≥ f (15) ≥ ldots ldots . f (99) ldots . . text (1) 3,9,15, ldots . .99 ⇒ 17 numbers for condition one we have 50 C 17 × 1 way rest 33 elements 33 ! = 50 C 17 × 33 ! = 50 C 33 × 33 ! = 50 P 33 .

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

$^{50} P_{17}$

$^{50} P_{33}$

$33! \times 17!$

$\frac{50 !}{2}$

Q. The number of bijective functions f:{1,3,5, 7,……..99}→{2,4,6,8,……..,100}, such that f(3)≥f(9)≥f(15)≥f(21)≥…..f(99), is _____

50P17​

50P33​

33!×17!

250 ! ​

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​.

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

$^{50} P_{17}$

$^{50} P_{33}$

$33! \times 17!$

$\frac{50 !}{2}$