JEE Advanced Mathematics Question (2026) — Solution

Step-by-step solution

An equivalence relation on a set S corresponds to a partition of S into disjoint equivalence classes. Let the sizes of these equivalence classes be a_1, a_2, , a_k. The number of elements in the equivalence relation R is given by the sum of the squares of the sizes of its equivalence classes: _ i=1 ^k a_i^2 = 42 Since the total number of elements in S is 10, we also have: _ i=1 ^k a_i = 10 We need to find all possible partitions of 10 such that the sum of their squares is 42. Let's check the possible sizes of the largest equivalence class, a_1: Case 1: a_1 = 6 a_1^2 = 36. The remaining sum of squares is 42 - 36 = 6, and the remaining sum of elements is 10 - 6 = 4. The only way to partition 4 such that the sum of squares is 6 is 2, 1, 1 (since 2^2 + 1^2 + 1^2 = 6). Thus, one valid partition is \ 6, 2, 1, 1\ . Case 2: a_1 = 5 a_1^2 = 25. The remaining sum of squares is 42 - 25 = 17, and the remaining sum of elements is 10 - 5 = 5. The only way to partition 5 such that the sum of squares is 17 is 4, 1 (since 4^2 + 1^2 = 17). Thus, another valid partition is \ 5, 4, 1\ . Case 3: a_1 4 The maximum possible sum of squares would be for the partition \ 4, 4, 2\ , which gives 4^2 + 4^2 + 2^2 = 36 Now, we calculate the number of ways to form these partitions from the 10 elements of S. For the partition \ 6, 2, 1, 1\ : The number of ways to divide 10 elements into groups of sizes 6, 2, 1, 1 is: 10! 6! 2! 1! 1! 2! = 3628800 720 2 1 1 2 = 3628800 2880 = 1260 (Note: We divide by 2! because there are two groups of identical size 1). For the partition \ 5, 4, 1\ : The number of ways to divide 10 elements into groups of sizes 5, 4, 1 is: 10! 5! 4! 1! = 3628800 120 24 1 = 3628800 2880 = 1260 Total number of equivalence relations in X is: 1260 + 1260 = 2520 Answer: 2520