JEE Main 2020 · 8 January · morning · Single correct · +4 / −1

If a, b and c are the greatest value of ¹⁹Cₚ, ²⁰C q and ²¹Cᵣ respectively, then :

  1. (a) a/11 = b/22 = c/21
  2. (b) a/10 = b/22 = c/21
  3. (c) a/10 = b/11 = c/42
  4. (d) a/11 = b/22 = c/42
Answer and solution

Answer (d) a/11 = b/22 = c/42

Solution

Maths sir · 4 steps

Pick the middle term each time, then link the three using ⁿCᵣ = n/r ⁿ⁻¹Cᵣ₋₁.

  1. Greatest value. ⁿCᵣ is largest at the middle. For even n that is r = n/2. For odd n there are two equal middle values, at r = (n-1)/2 and (n+1)/2. So a = ¹⁹C₉, b = ²⁰C₁₀ and c = ²¹C₁₁.

    ⁿCᵣ = n/r ⁿ⁻¹Cᵣ₋₁

  2. Link b to a. b = ²⁰C₁₀ = 20/10 ¹⁹C₉ = 2a, so b/22 = 2a/22 = a/11.
  3. Link c to b. c = ²¹C₁₁ = 21/11 ²⁰C₁₀ = 21/11 × 2a = 42a/11, so c/42 = a/11. As a check, a = 92378, b = 184756 and c = 352716 all give 8398.
  4. So. a/11 = b/22 = c/42, which is option (d).

The idea to keep. Neighbouring middle binomial coefficients are linked by ⁿCᵣ = n/r ⁿ⁻¹Cᵣ₋₁. Use that ratio instead of computing big factorials.

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.