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I wanna know the answer of this problem and solve it:

$$\binom{10}{0}^2+\binom{10}{1}^2+\binom{10}{2}^2+\cdots+\binom{10}{10}^2 = \ ?$$

Can you help me? Thanks in advance.

Lorenzo B.
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user3832258
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  • Are you asking for a general solution for $\sum_{k=0}^n \binom{n}{k}^2$ or only for the special case where $n=10$? – mrtaurho Jul 23 '18 at 08:08

1 Answers1

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$$\binom{2n}{n} = \sum_{k=0}^n \binom{n}{k} \binom{n}{n-k} = \sum_{k=0}^n \binom{n}{k}^2$$ I think it will help

D F
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  • I want the Prove of this way. Can you write it? – user3832258 Jul 23 '18 at 08:25
  • Also I am not sure what to add to this answer @user3832258. Just plug in $n=10$ and you have you solution. – mrtaurho Jul 23 '18 at 08:31
  • Thanks a lot. Can you guide me in proving this math formula? – user3832258 Jul 23 '18 at 08:34
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    @user3832258 So, we want to pick $n$ elements out of $2n$, each sample can be presented as: $k$ elements out of the first $n$ elements and $n-k$ elements out of the second $n$ elements $\ \forall \ k = 0, 1, \dots, n$, thus, due to the multiplication rule $\binom{2n}{n} = \sum_{k=0}^n \binom{n}{k} \binom{n}{n-k}$, but $\binom{n}{k} = \binom{n}{n-k}$ hence $\binom{2n}{n} = \sum_{k=0}^n \binom{n}{k}^2$ – D F Jul 23 '18 at 08:34