JEE Main 2020 · 7 January · evening · Single correct · +4 / −1

The number of ordered pairs (r, k) for which
6.³⁵Cᵣ = (k² - 3). ³⁶C r + 1 , where k is an integer, is :

  1. (a) 6
  2. (b) 3
  3. (c) 2
  4. (d) 4
Answer and solution

Answer (d) 4

Solution

Maths sir · 3 steps

Turn ³⁶Cᵣ₊₁ into ³⁵Cᵣ, and r and k get tied by one simple equation.

  1. Shrink the coefficient. Use ⁿCᵣ = n/r ⁿ⁻¹Cᵣ₋₁ with n = 36. That gives ³⁶Cᵣ₊₁ = 36/(r+1) ³⁵Cᵣ.

    6 ³⁵Cᵣ = (k²-3) 36/(r+1) ³⁵Cᵣ ⇒ r+1 = 6(k²-3)

  2. Limit the values. ³⁵Cᵣ needs 0 ≤ r ≤ 35, so 1 ≤ r+1 ≤ 36. Then 1 ≤ 6(k²-3) ≤ 36, which gives 3 < k² ≤ 9. The only perfect squares here are k² = 4 and k² = 9.
    k²r + 1 = 6(k²-3)rPairs (r, k)
    465(5, 2), (5, -2)
    93635(35, 3), (35, -3)
  3. So. Each allowed k² gives two integer values of k, so there are 4 ordered pairs, option (d).

The idea to keep. When two binomial coefficients differ by one in both n and r, rewrite one as a multiple of the other. Then use 0 ≤ r ≤ n to box in the integer values.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Binomial Theorem

Asked about 200 times in JEE Main since 2020. This question is from 2020.