Let ( n; k ) denotes ⁿCₖ and [ n; k ] = { ( n; k ),, if 0 ≤ k ≤ n; 0,, otherwise
If Aₖ = Σᵢ₌₀⁹ ( 9; i )[ 12; 12 - k + i ] + Σᵢ₌₀⁸ ( 8; i )[ 13; 13 - k + i ] and A₄ - A₃ = 190 p, then p is equal to :

Solution

Maths sir · 4 steps

Flip C(12, 12-k+i) to C(12, k-i) and each sum becomes Vandermonde, giving C(21, k).

1. Use symmetry. Since C(n, r) = C(n, n-r), the bracket [12; 12-k+i] equals C(12, k-i). It is 0 exactly when k-i falls outside 0 to 12, which is what the bracket definition says. In the same way, [13; 13-k+i] = C(13, k-i).
2. Apply Vandermonde. The sum Σᵢ C(m, i)C(n, k-i) counts ways to choose k items from m+n, so it equals C(m+n, k). Both sums have m + n = 21: 9 + 12 and 8 + 13.

   Aₖ = C(21, k) + C(21, k) = 2C(21, k)

3. Put in numbers. C(21, 4) = 5985 and C(21, 3) = 1330. So A₄ - A₃ = 2(5985 - 1330) = 2 × 4655 = 9310.
4. So. 190p = 9310, which gives p = 9310/190 = 49.

The common slip: 24.5. Counting only one of the two sums leaves Aₖ = C(21, k) and gives p = 4655/190 = 24.5. Both sums equal C(21, k), so Aₖ is twice that.

The idea to keep. When a sum multiplies two binomial coefficients and their lower indices add to a fixed k, think Vandermonde: Σ C(m, i)C(n, k-i) = C(m+n, k). Use symmetry first to make the lower indices add up.

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