JEE Main 2022 · 25 Jul · Shift 2 · Mathematics Q3

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MathematicsSingle correct+4 / -1Question 3 of 90

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

{}^{50}P_{17}

{}^{50}P_{33}

33!\times17!

\dfrac{50!}2

Answer and worked solution

Official answer: B

Fixed order. The 17 inputs $3,9,\dots,99$ take 17 values in decreasing order: ${}^{50}C_{17}$ ways. The other 33 inputs are a bijection onto the rest: $33!$ .

Total ${}^{50}C_{17}\,33!={}^{50}P_{33}$ .

(B)

Topics:Combinations Types of functions

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JEE Main 2022 · 25 Jul · Shift 2, Mathematics question 3