Vandermonde Identity

最新推荐文章于 2023-05-20 17:52:56 发布

DerekHJH

最新推荐文章于 2023-05-20 17:52:56 发布

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分类专栏：数学文章标签：组合数范德蒙不等式

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本文介绍了范德蒙德恒等式的两种证明方法：直观解释涉及从组合学角度理解从n+m个学生中选择k个的场景；数学证明则通过赋值和数学归纳法展示其正确性。此外，还讨论了该恒等式在特定情况下的应用，例如当k=m时的特殊形式，并强调了参数的灵活性在证明和应用中的价值。

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$\binom{n+m}{k}=\sum_{i=0}^{k}\binom{n}{i}\binom{m}{k-i}$ is calld Vandermonde Identity. According to Wikipedia and some other 优快云 blogs, I breifly summarize the way of proving this identity as follows:

The first version of proof is quite intuitive. We explain this identity as that there are n+m students in a class, and we want to pick k of them out. This intuition actually tells us that whenwe try to tackle some problems related to binomial coefficient, give it a vivid context, which will help us understand the problem better.

The second verion is more mathematical.

$\sum_{k=0}^{n+m}\binom{n+m}{k}x^k=(1+x)^{n+m}=(1+x)^n(1+x)^m=(\sum_{i=0}^{n}\binom{n}{i}x^i)(\sum_{j=0}^{m}\binom{m}{j}x^j)=\sum_{k=0}^{n+m}\sum_{i=0}^{k}\binom{n}{i}\binom{m}{k-i}x^k$ Let x=1, and we have

$\sum_{k=0}^{n+m}\binom{n+m}{k}=\sum_{k=0}^{n+m}\sum_{i=0}^{k}\binom{n}{i}\binom{m}{k-i}$ Thus we guess

$\binom{n+m}{k}=\sum_{i=0}^{k}\binom{n}{i}\binom{m}{k-i}$ . And the rest of the thing is simply to prove this by using the method of mathematical indeuction which is trivial and we skip this part.

Now we talk a little more about this identity.

We assume that m<=n, and by replacing k=m, we have $\binom{n+m}{m}=\sum_{i=0}^{m}\binom{n}{i}\binom{m}{m-i}=\sum_{i=0}^{m}\binom{n}{i}\binom{m}{i}$ , which is another beautiful identity we should pay attention to.