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$a , b, c$ are positive integers such that ${a}\geq{b+c}$.

Prove that $$\sum_{i=0}^{b}{\binom{a}{i+c}\binom{b}{i}}=\binom{a+b}{b+c}$$

acat3
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  • Share your own attempts at this problem. People will want to help you more if you do so – Aniket Gupta Feb 23 '20 at 15:26
  • This is a problem i created when i was solving another combinatorics problem. I already knew the answer. – acat3 Feb 23 '20 at 15:27
  • Better start at $i=0$. – RobPratt Feb 23 '20 at 15:42
  • See also https://math.stackexchange.com/questions/1659106/prove-that-sum-limits-k-0m-binommk-binomnrk-binommnmr-using – Robert Z Feb 23 '20 at 15:46
  • Ah, the beauty of combinatorics, there are always multiple solutions to a problem. I guess it is unfair to expect people to give a particular answer to the problem. I will share my methods and just close the question. – acat3 Feb 23 '20 at 15:55

3 Answers3

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$\binom{a}{i+c}$ is the coefficient of $x^{i+c}$ from $(x+1)^{a}$ while $\binom{b}{i}$ is the coefficient of $\frac{1}{x^{i}}$ from $(\frac{1}{x}+1)^{b}$.

$\sum_{i=0}^{b}\binom{a}{i+c}\binom{b}{i}$ is the coefficient of $x^{c}$ from $(x+1)^{a}(\frac{1}{x}+1)^b=\frac{(x+1)^{a+b}}{x^{b}}$ which is $\binom{a+b}{b+c}$.

acat3
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1

Write the first term on the left side as $\binom{a}{ a - c - i}$ and the term in the right side as $\binom{a + b}{ a - c}$. (Hypergeometric distribution?)

Qurultay
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By Vandermonde, $$\sum_{i=0}^{b}{\binom{a}{i+c}\binom{b}{i}}=\sum_{i=0}^{b}{\binom{a}{a-c-i}\binom{b}{i}}=\binom{a+b}{a-c}=\binom{a+b}{b+c}$$

RobPratt
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