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

(B) ${ }^{50} \mathrm{P}_{33}$

(A) ${ }^{50} \mathrm{P}_{17}$

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Permutations and Combinations

The number of bijective functions

f : {1, 3, 5, 7, \dots, 99} \to {2, 4, 6, 8, \dots, 100}

, such that

f (3) \geq f (9) \geq f (15) \geq f (21) \geq \dots \geq f (99)

, is

Answer: (B)

The domain and codomain each contain

50

elements. The constrained arguments

3, 9, 15, \dots, 99

form an arithmetic progression with common difference

6

, so there are

\frac{99 - 3}{6} + 1 = 17

of them. Since

f

is a bijection every image is distinct, so the chain

f (3) \geq f (9) \geq \dots \geq f (99)

is in fact strictly decreasing. Choosing the

17

even images for these positions can be done in

(17 50)

ways, and for each choice the decreasing order fixes their assignment uniquely. The remaining

33

odd numbers then map bijectively onto the remaining

33

even numbers in

33!

ways. The total is

(17 50) \cdot 33! = \frac{50 !}{17 ! 33 !} \cdot 33! = \frac{50 !}{17 !} =^{50} P_{33}

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option