JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given A = \ 1, 2, 3, 4, 5, 6\ . Since f: A A is a one-one function, f(x) takes distinct values from A for each x A. We are given the condition f(2) + f(3) = 5. The possible pairs for (f(2), f(3)) from the set A are: (1, 4), (2, 3), (3, 2), (4, 1) Notice that for all these pairs, the condition f(3) 4 is automatically satisfied. We also have the condition f(1) 3. Since f is one-one, f(1) cannot take the values already assigned to f(2) and f(3). Let us find the number of choices for f(1) in each case: Case 1: (f(2), f(3)) = (1, 4) f(1) 3 and f(1) \ 1, 4\ f(1) \ 3, 5, 6\ . (3 choices) Case 2: (f(2), f(3)) = (2, 3) f(1) 3 and f(1) \ 2, 3\ f(1) \ 4, 5, 6\ . (3 choices) Case 3: (f(2), f(3)) = (3, 2) f(1) 3 and f(1) \ 3, 2\ f(1) \ 4, 5, 6\ . (3 choices) Case 4: (f(2), f(3)) = (4, 1) f(1) 3 and f(1) \ 4, 1\ f(1) \ 3, 5, 6\ . (3 choices) In each of the 4 cases, there are exactly 3 choices for f(1). After assigning values to f(1), f(2), and f(3), there are exactly 3 elements left in the codomain to be assigned to f(4), f(5), and f(6). Since f is one-one, these can be assigned in 3! = 6 ways. Total number of such functions = (Number of pairs) (Choices for f(1)) (Arrangements for remaining elements) Total = 4 3 6 = 72 Answer: 72