If n+4 n+1 - ^ n+3 C_n=15(n+2) , then n=

Step-by-step solution

Applying the symmetry property ^N C_R = ^N C_ N-R , we rewrite: n+4 n+1 = ^ n+4 C₃ ^ n+3 C_n = ^ n+3 C₃ The original equation becomes: ^ n+4 C₃ - ^ n+3 C₃ = 15(n+2) Using the identity ^N C_R - ^ N-1 C_R = N-1 R-1 with N = n+4 and R = 3 : ^ n+4 C₃ - ^ n+3 C₃ = ^ n+3 C₂ Thus, the equation simplifies to: ^ n+3 C₂ = 15(n+2) Expanding the combination: ^ n+3 C₂ = (n+3)(n+2) 2 Substituting back: (n+3)(n+2) 2 = 15(n+2) Since n 0 for valid combinations, we can divide both sides by (n+2) : n+3 2 = 15 Solving yields n = 27 .