This comes from the binomial expansion of (X+y)^n, where X^k y^(n-k) has coefficient (n choose k). This is since you "choose" X in K of the (X+y)'s (in the Bezi

This comes from the binomial expansion of (X+y)^n, where X^k y^(n-k) has coefficient (n choose k). This is since you "choose" X in K of the (X+y)'s (in the Bezier case it's kind of writing (t+(1-t))^n, the abcd are to not make it equal to 1). The choice function is also exactly the same as the pascal triangle!