The value of Σᵣ₌₀²² ²²Cᵣ²³Cᵣ is

Solution

Maths sir · 4 steps

Flip one coefficient so the lower indices add to a constant, then read it as one coefficient of a product.

1. Flip one index. Use ²³Cᵣ = ²³C₂₃₋ᵣ. Each term becomes ²²Cᵣ · ²³C₂₃₋ᵣ, and now the lower indices always add to r + (23 - r) = 23.
2. Read as coefficient. ²²Cᵣ is the coefficient of xʳ in (1+x)²², and ²³C₂₃₋ᵣ is the coefficient of x²³⁻ʳ in (1+x)²³. Each product is one way to build x²³, so the sum is the coefficient of x²³ in (1+x)²²(1+x)²³.

   Σᵣ₌₀²² ²²Cᵣ ²³C₂₃₋ᵣ = coefficient of x²³ in (1+x)⁴⁵ = ⁴⁵C₂₃

3. Check the limits. The pairing would also allow r = 23, but that term has ²²C₂₃ = 0. So stopping at r = 22 misses nothing.
4. So. The sum equals ⁴⁵C₂₃, which is option (b).

The idea to keep. For a sum of products of two binomial coefficients, make the lower indices add to a constant k. The sum is then the coefficient of xᵏ in the product, and the upper indices add.

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