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Numerically calculating the sum of the squares of the $m$th row of Pascal's triangle, I found that for at least the first $10$ or so cases $$\sum_{i=0}^m \binom{m}{i}^2=\frac{(4m-2)!!!!}{m!}$$ Where $(4k-2)!!!!=(4k-2)(4k-6)(4k-10)\cdots 6\cdot2=2^m(2m-1)!!$.

Of course the problem reduces to, using the well-known formula,

$$\frac{(2m)!}{(m!)^2}=\frac{2^m(2m-1)!!}{m!}$$ $$(2m-1)!!=\frac{(2m)!}{m!2^m}$$

Is there a direct proof for the above equation, or (second-best) is there a proof by proving my initial guess correct?

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    $\begingroup$ $m! 2^m = 2 \cdot 4 \cdot ... \cdot (2m)$. Is it clear now? $\endgroup$ – Qiaochu Yuan Jun 19 '13 at 22:10
  • $\begingroup$ @QiaochuYuan Yes, silly oversight by me, thanks. $\endgroup$ – Meow Jun 19 '13 at 22:11
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The left side is the product of all the odd numbers from $1$ through $2m-1$. The numerator on the right is the product of all the numbers $1$ through $2m$. If you multiply each factor of $m!$ by one of the $2$'s you divide out all the evens from the numerator.

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