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!
Comments
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!