JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

Given equation: ^ 36 C_ r+1 = 6 (^ 35 C_r ) k^2 - 3 Using the property ^ n C_r = n r ^ n-1 C_ r-1 , we can write: ^ 36 C_ r+1 = 36 r+1 ^ 35 C_r Substituting this into the given equation: 36 r+1 ^ 35 C_r = 6 k^2 - 3 ^ 35 C_r For ^ 35 C_r to be defined, r must be an integer such that 0 r 35. Thus, ^ 35 C_r 0. Dividing both sides by ^ 35 C_r: 36 r+1 = 6 k^2 - 3 6 r+1 = 1 k^2 - 3 k^2 - 3 = r+1 6 k^2 = 3 + r+1 6 For k Z , k^2 must be a perfect square integer. This requires r+1 to be a multiple of 6. Since 0 r 35, we have 1 r+1 36. Let r+1 = 6m, where m \ 1, 2, 3, 4, 5, 6\ . Then k^2 = 3 + m. Checking the possible values of m for which k^2 is a perfect square: For m = 1, k^2 = 4 k = 2. (Here r+1 = 6 r = 5) For m = 2, k^2 = 5 (not a perfect square). For m = 3, k^2 = 6 (not a perfect square). For m = 4, k^2 = 7 (not a perfect square). For m = 5, k^2 = 8 (not a perfect square). For m = 6, k^2 = 9 k = 3. (Here r+1 = 36 r = 35) The possible pairs (r, k) are (5, 2), (5, -2), (35, 3), and (35, -3). Therefore, the number of elements in the set S is 4. Answer: 4