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Vandermondes Identity

  1. Jul 28, 2007 #1
    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:

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

    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. jcsd
  3. Aug 3, 2007 #2
    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
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