text: is this true? that nc(n-b) + (n-1)c(n-b-1) + (n-2)c(n-b-2) + ... + bc0 = (n+1)c(n+b) where 0 (</=) b (</=) n  i've formulated this question from the pascals triangle... you will notice that the sum of #s on 1 diagonal downwards to the left from any 1 is equal to the number to the right of the number next to the last number in your sum!! (just translated it mathematically!!^_^ it's not quite right.  you can tweak the formula to make it work.  yours:  nc(n-b) + (n-1)c(n-b-1) + (n-2)c(n-b-2) + ... + bc0 = (n+1)c(n+b) where  0 (</=) b (</=) n  actual: nc(n-b) + (n-1)c(n-b-1) + (n-2)c(n-b-2)+...+(n-b)c0= (n+1)cb  try that and see if it looks ok