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All positive integers $m_1, m_2, n$ satisfy:

$$ \sum_{k=0}^n {m_1 \choose k}{m_2 \choose n-k} = {m_1 + m_2 \choose n} $$

Prove using the binomial theorem and the fact that $(1+x)^{m_1}(1+x)^{m_2} = (1+x)^{m_1+m_2}$

To my understanding, $ (1+x)^{m_1} = \sum_{k=0}^n {m_1 \choose k}x^k $. Continuing with this, wouldn't the left side be a summation of a summation? Where does $ x^k$ go?

Using the given, I imagine that $(1+x)^{m_1+m_2}$ could be used to retrieve the right side but then I don't know where sigma goes or how it becomes "choose n"

Any hints would be appreciated.

Edit: I feel really dumb. Thanks everyone!

user194351
  • 31

4 Answers4

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Assume $m_1\le m_2$ without loss of generality. Suppose $n\le m_1$. You know that, by binomial theorem for any integer $r$,

$$ (1+x)^m=\sum_{k=0}^{m} {m \choose k} x^k $$

So $$ (1+x)^{m_1}(1+x)^{m_2}=\left(\sum_{i=0}^{m_1} {m_1 \choose i}x^i\right) \left(\sum_{j=0}^{m_2} {m_2 \choose j} x^j\right)=\sum_{k=0}^{m_1+m_2} {m_1+m_2 \choose k} x^k=(1+x)^{m_1+m_2} $$

The two expressions must equal for all $x$, so the coefficient must match. For instance, what is coefficient of $x^n $ (when $k=n$) in RHS of above equation, and what is coefficient of $x^n$ (when $i+j=n$)?

user160738
  • 4,160
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Working out: $$\left(1+x\right)^{m_{1}}\left(1+x\right)^{m_{2}}=\left[\sum_{r=0}^{m_{1}}\binom{m_{1}}{r}x^{r}\right].\left[\sum_{s=0}^{m_{2}}\binom{m_{2}}{s}x^{s}\right]$$ you will get a polynomial where: $$\sum_{k=0}^{n}\binom{m_{1}}{k}\binom{m_{2}}{n-k}$$ shows up as coefficient of $x^{n}$.

Next to that: $$\left(1+x\right)^{m_{1}}\left(1+x\right)^{m_{2}}=\left(1+x\right)^{m_{1}+m_{2}}=\sum_{t=0}^{m_{1}+m_{2}}\binom{m_{1}+m_{2}}{t}x^{t}$$ where: $$\binom{m_{1}+m_{2}}{n}$$ shows up as coefficient of $x^{n}$.

drhab
  • 151,093
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Hint: compering coefficients of both side polynomials.

H.Kakuhama
  • 109
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Here's a combinatorial explanation: you need to select $n$ from $m_1 +m_2$ items, this is $\binom{m_1+m_2}{n}$. But this is the same as picking $0$ items from $m_1$ and $n$ from $m_2$ or 1 item from $m_1$ and $n-1$ from $m_2$, etc. Hence $$ \sum_{k=0}^{n} \binom{m_1}{k} \binom{m_2}{n-k} = \binom{m_1 +m_2}{n} $$

No algebra was used here.

Alex
  • 19,262