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One of the most important and basic identities in binomial summations in Vandermonde's identity which states :

$$\sum_{k=0}^{r}\binom{m}{k}\binom{n}{r-k}=\binom{m+n}{r}$$

The identity can be generalized to the case when more than two binomial coefficients are multiplied.

Setting $m,r \mapsto n$ follows:

$$\sum_{k=0}^{n}\binom{n}{k}\binom{n}{n-k}=\binom{2n}{n}=\sum_{k=0}^{n}\binom{n}{k}^2$$

Which is very well-known.

One of the questions related to this summation is :

$$\sum_{k=0}^{n}\binom{n}{k}^{4}=\sum_{k=0}^{n}\binom{\color{red}{n}}{\color{blue}{k}}\binom{\color{red}{n}}{\color{blue}{k}}\binom{\color{red}{n}}{\color{blue}{n-k}}\binom{\color{red}{n}}{\color{blue}{n-k}}$$

I thought that the answer would be $\binom{4n}{2n}$, but this is not right in general and I don't know why, since the sum of the red and blue parts is independent of the index (is a fixed number),so that's why I used the convolution (It would be highly appreciated if someone explain where was I wrong).

My question is : Does there exist any closed formula for :

$$\sum_{k=0}^{n}\binom{n}{k}^{r}$$

Where $r \in \mathbb N$.

One of the identities related to this question is Dixon’s Identity which has been mentioned here (page 3).

wolframalpha does not give a closed form.

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    $\begingroup$ I don't have an answer to your main question right now, but the reason your $\binom nk^4$ sum does not work is that the sum does not range over all terms $\binom na \binom nb \binom nc \binom nd$ where $a+b+c+d=2n$, only those where $a=b$ and $c=d$. $\endgroup$ – Misha Lavrov Mar 15 '20 at 18:56
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    $\begingroup$ Try $r=3$ first. This should give a sufficient disappointment. $\endgroup$ – metamorphy Mar 15 '20 at 19:05
  • $\begingroup$ @ metamorphy, you are totally right $\endgroup$ – user715522 Mar 15 '20 at 19:06
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    $\begingroup$ I didn't think it through and I get the problem now. So basically $ (^1_0)^4+(^1_1)^4 =2 \ne 4=C(4,2)$. And the reason for that is if we consider $ k_1+k_2+k_3+k_4=2$ where $k_i \in {0,1}$, we have $\frac{4!}{2!2!}$ solutions and thus six ways to get the coefficient of $x^2$ from $(1+x)^4$ $\endgroup$ – Mathsmerizing Mar 15 '20 at 19:08
  • $\begingroup$ @ metamorphy , you are right, and I guess you meant $\binom{4}{2}=6$ $\endgroup$ – user715522 Mar 15 '20 at 19:18

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