Vandermonde's Identity: Exploring a Special Case with Binomial Coefficients

[ehrenfest]

Homework Statement

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:

[tex] \sum_{k=0}^r \binom{m}{r-k} * \binom{n}{k} = \binom{m+n}{r} [/tex]

Homework Equations

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...

 

[matness]

A late reply

You possibly end up with [tex] \binom{m+1}{a+1} = \sum_{k=0}^{a+1}
\binom{m}{a+1-k} * \binom{1}{k} [/tex]

I am not sure but , I guess [tex] \binom{1}{k} =0[/tex] if k>1 by the definition of combination
Because the number of ways picking e.g. 3 elements from a set with 1 element must be zero.
So then you can get what you want only summing two terms as you said
n

 

[Architeuthis Dux]

Yes, the identity C(m, a) + C(m,a+1) = C(m+1,a+1) is a special case of Vandermonde's identity. It can be rewritten as:

C(m, a) = C(m, a+1) + C(m+1, a+1)

This is equivalent to:

\binom{m}{a} = \binom{m+1}{a} + \binom{m+1}{a+1}

which is a special case of Vandermonde's identity with m+1 and a+1 as the parameters. The original identity is a specific example of this more general form, where n=1 and r=a. So, while it may not seem obvious at first glance, the identity is indeed a special case of Vandermonde's identity.