If for some m, n ; ⁶ Cₘ+2( ⁶ Cₘ₊₁)+ ⁶ Cₘ₊₂> ⁸ C₃ and ⁿ⁻¹ P₃: ⁿ P₄=1: 8, then ⁿ Pₘ₊₁+ ⁿ⁺¹ Cₘ is equal to

Solution

Maths sir · 4 steps

The 1, 2, 1 pattern is Pascal's rule used twice, so the left side folds into a single ⁸C.

1. Pascal twice. Split the middle term into two copies and pair them up. Use ⁿCᵣ + ⁿCᵣ₊₁ = ⁿ⁺¹Cᵣ₊₁ to combine each pair, then use it once more on the result.

   (⁶Cₘ + ⁶Cₘ₊₁) + (⁶Cₘ₊₁ + ⁶Cₘ₊₂) = ⁷Cₘ₊₁ + ⁷Cₘ₊₂ = ⁸Cₘ₊₂

2. Find m. We need ⁸Cₘ₊₂ > ⁸C₃ = 56. The values ⁸Cᵣ run 1, 8, 28, 56, 70, 56, 28, 8, 1, and only the middle one, ⁸C₄ = 70, beats 56. So m + 2 = 4, which gives m = 2.
3. Find n. ⁿP₄ = n(n-1)(n-2)(n-3) = n × ⁿ⁻¹P₃. So ⁿ⁻¹P₃/ⁿP₄ = 1/n = 1/8, which gives n = 8.
4. So. ⁿPₘ₊₁ + ⁿ⁺¹Cₘ = ⁸P₃ + ⁹C₂ = 336 + 36 = 372, which is option (c).

The idea to keep. When binomial coefficients appear with weights 1, 2, 1 on consecutive r, apply Pascal's rule twice: ⁿCᵣ + 2 ⁿCᵣ₊₁ + ⁿCᵣ₊₂ = ⁿ⁺²Cᵣ₊₂.

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