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My attempt $$\sum_{r=0}^{n} r \times \frac nr \left[\binom {n-1}{r-1} \right]^2$$ $$=n\sum _{r=0}^{n} \left [\binom {n-1}{r-1}\right]^2$$

The square on the binomial coefficient is throwing me off, otherwise it would have been easier to proceed. I don’t know what to do with that square.

Edit: As pointed out in the comment, i managed to find out $$\sum_{r=0}^n (\binom nr )^2 = \binom {2n}{n}$$

But using this in the above expression isn’t giving me the right answer. Am I supposed to make a modification to the limits in the sigma function?

Aditya
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  • Hint: binomial coefficients have a symmetry: $\binom{n}{r} = \binom{n}{n-r}$. Try rewriting the sum using a product of both types in each term. – Sammy Black Feb 18 '21 at 04:13
  • $r\binom{n}r^2=n\binom{n-1}{r-1}\binom{n}r$, not $n\binom{n-1}{r-1}^2$. And you can usefully rewrite that as $n\binom{n-1}{r-1}\binom{n}{n-r}$ – Brian M. Scott Feb 18 '21 at 04:27
  • @Aditya: I got it by doing the algebra correctly: $$\begin{align}r\binom{n}r^2&=n\cdot\frac{r}n\cdot\binom{n}r\cdot\binom{n}r\&=n\binom{n-1}{r-1}\binom{n}r\end{align}$$ – Brian M. Scott Feb 18 '21 at 04:34
  • Be careful of the sum limits. Better omit $r=0$. – herb steinberg Feb 18 '21 at 04:38
  • @BrianM.Scott so how do I sum that expression? I know the sum for $\binom {n}{r-1} \binom nr = \binom {2n} {n-1}$ but not with $n-1$ – Aditya Feb 18 '21 at 04:40
  • Given $n$ google employees and $n$ apple employees, pick a combined team of $n$ with the captain from google – Empy2 Feb 18 '21 at 04:49
  • @Empy2 is it $\binom {2n-1} {n-1}$? And how is that related to this problem? – Aditya Feb 18 '21 at 04:52
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    @Aditya: It should be clear that you just need to compute $\sum_r\binom{n-1}{r-1}\binom{n}r$ and multiply by $n$. $\binom{n-1}{r-1}=\binom{n-1}{n-r}$, so you can compute $\sum_r\binom{n}r\binom{n-1}{n-r}$ instead. Imagine that you have $n$ women and $n-1$ men, and you want to pick a team of $n$ people from this pool of $2n-1$ people. $\binom{n}r\binom{n-1}{n-r}$ is the number of ways to pick a team with $r$ women. Summing over $r$ gives you the number of possible teams, but you can calculate that directly: how many teams of $n$ can be formed from a pool of $2n-1$ people? – Brian M. Scott Feb 18 '21 at 05:01
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    Does this answer your question? (also (1), (2), more general (3)...) – metamorphy Feb 18 '21 at 06:42
  • Pick $r$ from google, omit $r$ of the $n$ apple, then make one of the $r$ googles captain. Or pick $n$ from the combined $2n$, pick a captain from them, anf half the time the captain turns out to be from google – Empy2 Feb 18 '21 at 11:57

1 Answers1

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Symmetry is the key. We obtain \begin{align*} \sum_{r=0}^nr\binom{n}{r}^2&=\sum_{r=0}^n(n-r)\binom{n}{r}^2\tag{1}\\ &=n\sum_{r=0}^n\binom{n}{r}^2-\sum_{r=0}^nr\binom{n}{r}^2\\ \end{align*} where we changed in the right-hand sum of (1) the order of summation $r\to n-r$. We already know according to OPs calculation: \begin{align*} \sum_{r=0}^n\binom{n}{r}^2=\binom{2n}{n}\tag{2} \end{align*}

From (1) and (2) we obtain \begin{align*} \color{blue}{\sum_{r=0}^nr\binom{n}{r}^2}&=\frac{n}{2}\sum_{r=0}^n\binom{n}{r}^2 \color{blue}{=\frac{n}{2}\binom{2n}{n}} \end{align*} and the claim follows.

Markus Scheuer
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  • Shouldn’t the limits also change if you are replacing $r\to n-r$? – Aditya Feb 19 '21 at 02:01
  • @Aditya: Let's consider for example the index range with $n=5$. We have the index sequence $(0,1,2,3,4,5)=(r){0\leq r\leq 5}$. Reverting the order of summation gives the index sequence $(5,4,3,2,1,0)=(5-r){0\leq r\leq 5}$. In both cases $0\leq r\leq 5$. – Markus Scheuer Feb 19 '21 at 10:04