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There was a post on this web site an hour ago asking for the sum of

\begin{equation*} \binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \end{equation*}

expressed as a single binomial coefficient. (Four choices were provided in the post.) The post seems to have been deleted. I think it is worth keeping on this web site.

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  • $\begingroup$ When I looked at it moments ago, the other question was not deleted (though it was just one vote away from being closed): math.stackexchange.com/questions/2195018 $\endgroup$
    – David K
    Mar 20, 2017 at 16:59
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    $\begingroup$ The older question was hard to find. For example, it never mentioned the word "binomial." I've added a couple of relevant tags to it, which might have helped. $\endgroup$
    – David K
    Mar 20, 2017 at 17:32

2 Answers 2

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Solution

For any positive integer $n$ and any nonnegative integer $r \leq n$,

\begin{equation*} \binom{n}{r} = \binom{n}{n - r} . \end{equation*} According to Vandermonde's Identity, \begin{align*} &\binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \\ &\qquad = \binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\ &\qquad = \sum_{r=0}^{49} \binom{50}{r} \binom{50}{49-r} \\ &\qquad = \binom{100}{49} . \end{align*}

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There is another proof for this problem that I am mentioning here :

Since :

$$ \binom{50}{0}\binom{50}{1} + \binom{50}{1}\binom{50}{2}+⋯+\binom{50}{49}\binom{50}{50} \\ \qquad = \binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\$$

Consider the product :

$$(1+x)^{50} \times (1+x)^{50}$$

Look for the coefficient of $x^{49}$ in this product, it will be calculated as :

$$ x^0 ~~\text{from first bracket} , x^{49} ~\text{from second bracket} = \binom{50}{0} \times \binom{50}{49}$$

$$ x^1 ~~\text{from first bracket} , x^{48} ~\text{from second bracket} = \binom{50}{1} \times \binom{50}{48}$$ $$ x^2 ~~\text{from first bracket} , x^{47} ~\text{from second bracket} = \binom{50}{2} \times \binom{50}{47} \\. \\. \\. \\.$$

$$ x^{49} ~~\text{from first bracket} , x^{0} \text{from second bracket} = \binom{50}{49} \times \binom{50}{0}$$

So, coefficient of $x^{49}$ in expansion of $(1+x)^{50}(1+x)^{50} =$

$$\binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0} \\$$

Which is coefficient of $x^{49}$ in expansion of $(1+x)^{100} =$

$$\binom{100}{49}$$

Thus :

$$\binom{50}{0}\binom{50}{49} + \binom{50}{1}\binom{50}{48}+⋯+\binom{50}{49}\binom{50}{0}=\binom{100}{49}$$

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  • $\begingroup$ That's really nice! $\endgroup$
    – Vincent
    Mar 20, 2017 at 18:29

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