Let X = (10C1)² + (10C2)² + 3(10C3)² + ... + 10(10C10)², where 10Cr, r ∈ {1, 2, ..., 10} denote binomial coefficients. Then, the value of 11430X is
Solution:
X = Σr=1¹⁰ r(ⁿCr)²; n = 10 ⇒ X = n Σr=1ⁿ r(ⁿCr)² ⇒ X = n Σr=1ⁿ r(ⁿCᵣ)(ⁿCₙ₋ᵣ) ⇒ X = n Σr=1ⁿ nCn₋ᵣ. nCr ⇒ X = n. ₂ₙ₋₁Cₙ₋₁; n = 10 ⇒ X = 10.₁₉C₉ ⇒ X/1430 = 11430/1430. ₁₉C₉ = 646