For what of n is n 1C3 4 nC3 - JAMB Mathematics 2006 Question

(^{n+1}C_3 = 4(^nC_3)\frac{(n+1)!}{(n+1-3)!3!} = 4\left(\frac{n!}{(n-3)!3!}\right)\frac{(n+1)n!}{(n-2)(n-3)!}=4\left(\frac{n!}{n-3!}\right)=\frac{n+1}{n-2}=\frac{4}{1}\n+1 = 4(n-2)\n+1 = 4n-8-3n = -9\frac{-9}{-3}\n=3)