JEE Main 2021 — Functions Question with Solution
From: JEE Main 2021 (Online) 22th July Evening Shift
Question
Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f : A A such that f(1) + f(2) = 3 f(3) is equal to
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Show full solutionCorrect answer: 720
Correct answer
720
Step-by-step explanation
f(1) + f(2) = 3 f(3)
f(1) + f(2) = 3 + f(3) = 3
The only possibility is : 0 + 1 + 2 = 3
Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.
So number of bijective functions.
= 720
f(1) + f(2) = 3 + f(3) = 3
The only possibility is : 0 + 1 + 2 = 3
Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.
So number of bijective functions.
= 720
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