Actual responses to actual questions from students of Math 344, Introduction to Combinatorics (Spring 2004).
Could you please provide some hints for how to solve part (b)?
Try first proving that C(n+r,r) C(n,r) = C(n+r,2r) C(2r,r).
The answer I came up with was -[n(1-1)(n-1)]. Could you confirm if this is indeed the correct answer?
Well, this is 0, the correct answer. I did this by evaluating the first derivative of x(1+x)n at x=-1, which gives you the desired ansewr (namely, zero) provided n is at least 2. Notice that if n=0 or n=1, the sum isn't zero.