If  n+2C3=n+3P2−20 then n is equal to:

16(n+2)(n+1)n=(n+3)(n+2)−20⇒     n3+3n2+2n=6n2+5n+6−120⇒     n3−3n2−28n+84=0⇒     (n−3)n2−28=0As n is a positive integer we get n=3