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Finding value of $$\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$$

Solution I tried:

$$(1+x)^{10}=1+\binom{10}{1}x+\binom{10}{2}x^2+\cdots +\binom{10}{10}x^{10}$$

$$(x+1)^{10}=x^{10}+\binom{10}{1}x^9+\binom{10}{2}x^7+\cdots +\binom{10}{10}$$

I did not find how do multiply terms such that i get my result. Help me

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closed as off-topic by Saad, Claude Leibovici, Tom-Tom, Namaste, Parcly Taxel May 24 '18 at 14:26

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    $\begingroup$ Hint: it's not so hard to do this by brute force. If you want a more elegant method, first prove combinatorially that this sum represents the number of ways to choose $5$ elements from $13$ objects. $\endgroup$ – lulu May 18 '18 at 18:25
  • $\begingroup$ lulu can you plz explain me $\endgroup$ – jacky May 18 '18 at 18:41
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    $\begingroup$ Suppose you are choosing $5$ objects out of $13$. list the choices as $n_1,\cdots, n_5$. Let $n_3=k+1$. Take it from there. $\endgroup$ – lulu May 18 '18 at 18:43
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To complete the method sketched in the comments.

The answer is $\binom {13}5$.

To see this, suppose we are choosing $5$ values from $\{1,\cdots, 13\}$. Order the choices as $n_1<\cdots <n_5$. Let $n_3=k+1$. Clearly $2≤k≤10$. Given a choice of such an $n_3$ we see that we are asked to choose $2$ values from $\{1,\cdots, k\}$ and $2$ from $\{k+2,\cdots, 13\}$. There are $\binom k2$ ways to do the former and $\binom {13-(k+2)+1}2=\binom {12-k}2$ ways to do the latter. Hence the number of choices with a given $n_3$ is the product $\binom {k}2\times \binom {12-k}2$. Summing over $k$ we see that $$\binom {13}5=\sum_{k=2}^{10}\binom {k}2\times \binom {12-k}2$$ as desired.

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  • $\begingroup$ Brilliant solution. You may want to mention that $n_3 = k+1$ can take values from $3$ to $13$ hence the sum runs on $k$ from $2$ to $12$. $\endgroup$ – feynhat May 18 '18 at 19:55
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    $\begingroup$ @feynhat $k+1$ can't be greater than $11$, since $13≥n_5>n_4>n_3=k+1$. Thus $3≤k+1≤11$ or, $2≤k≤10$. $\endgroup$ – lulu May 18 '18 at 19:59
  • $\begingroup$ In a way, you're both right; the $k\in\left\{11,\,12\right\}$ terms vanish because $12-k<2$. $\endgroup$ – J.G. May 18 '18 at 21:22
  • $\begingroup$ @lulu Yes. I meant to say "$n_3$ takes values from $3$ to $\textbf{11}$". $\endgroup$ – feynhat May 19 '18 at 3:22
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It is Chu–Vandermonde identity

$$\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}=\sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{4-2}={12+1\choose 4+1}=1287.$$

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Hint:

$$=\sum_{r=2}^{10}\dfrac{r(r-1)(12-r)(11-r)}4$$

Use the formulas in Geometric interpretation for sum of fourth powers

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$$S_{(10,2)} = \sum^{10}_{k=2}\binom{k}{2}\binom{12-k}{2}$$ You can generalize the question as: \begin{align} S_{(n,\,r)} &= \sum^{n}_{k=r}\binom{k}{r}\binom{(n+r)-k}{r}\\ \end{align}
Set $(n+r) = m\,$ and we get \begin{align} S_{(n,\,r)} &= \sum^{n}_{k=r}\binom{k}{r}\binom{m-k}{r}= \binom{m+1}{2r+1}\\ \end{align} (See Another form of the Chu–Vandermonde identity and this proof )


So

$$S_{(n,\,r)} = \binom{n+r+1}{2r+1}$$

In your case $$S_{(10,2)} = \binom{10+2+1}{2\cdot 2+1} = \binom{13}{5} = 1287$$

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