2
$\begingroup$

I have been trying to check that two things are equal for a while now, Mathematica appears to say that they are but I can't for the life of me figure out how to show it (without just saying 'computer says yes'), does anybody know a 'nice' way to show the following expression is true?

$$\binom{v + w + d - 1}{d} = \sum_{a=0}^d \binom{v+a-1}{a} \binom{w + d - a - 1}{d-a}$$

Improve this question
$\endgroup$
5
  • 7
    $\begingroup$ It is Vandermonde-Chu identity en.wikipedia.org/wiki/Vandermonde%27s_identity $\endgroup$
    – Fedor Petrov
    Commented Mar 8, 2016 at 18:58
  • $\begingroup$ Thank you! However, I can't quite get it to give me what I have above - am I being dense? If I use that formula directly I get $\binom{v+w+d-2}{d}$ instead of a $v + w + d - 1$ on the top and through direct calculation that isn't equal to the sum on the right. $\endgroup$
    – Jacksbabypig
    Commented Mar 8, 2016 at 19:37
  • 2
    $\begingroup$ Ah! You should at first use reflection $\binom{n+a-1}{a}=(-1)^a\binom{-n}{a}$. $\endgroup$
    – Fedor Petrov
    Commented Mar 8, 2016 at 20:05
  • $\begingroup$ See also Proving that ${x +y+n- 1 \choose n}= \sum_{k=0}^n{x+n-k-1 \choose n-k}{y+k-1 \choose k} $ and some other posts from math.SE returned in this Approach0 search. $\endgroup$
    – Martin Sleziak
    Commented Jan 6, 2017 at 9:33
  • $\begingroup$ The rhs is the coefficient of a Cauchy product of two binomial series, with exponent v and w resp, whence the identity. $\endgroup$
    – Pietro Majer
    Commented Mar 8, 2020 at 8:59

0

Reset to default

Browse other questions tagged .