# Vandermondes Identity

1. Jul 28, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
Is the identity C(m, a) + C(m,a+1) = C(m+1,a+1) (where C is the binomial coefficient function) a special case of Vandermonde's identity:

$$\sum_{k=0}^r \binom{m}{r-k} * \binom{n}{k} = \binom{m+n}{r}$$

2. Relevant equations

3. The attempt at a solution

n (or m) must equal 1 but r must also equal 1 because you only sum over two terms. But r obviously cannot be 1 since it is a + 1 on the right side. I am thinking that there some other forms of Vandermonde's identity which it may be a special case of but not this one? This form does not even make sense when r is greater than n and we need r to be general and n to be 1 I think...

Last edited: Jul 28, 2007
2. Aug 3, 2007

### matness

You possibly end up with $$\binom{m+1}{a+1} = \sum_{k=0}^{a+1} \binom{m}{a+1-k} * \binom{1}{k}$$
I am not sure but , I guess $$\binom{1}{k} =0$$ if k>1 by the definition of combination