2

If $\binom{n}{0} , \binom{n}{1} , \binom{n}{2} , \binom{n}{2} .... \binom{n}{n}$ denote the binomial coefficients in the expansion of ${(1+1)}^n$ then what is the difference between

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

and

$$\sum \sum \binom{n}{r}\binom{n}{s}$$

$r$ is greater than or equal to $0$ but it is less than $s$ and $s$ is less than or equal to $n$

How can we understand the differences in the meaning of the questions ?

Aditi
  • 1,349

3 Answers3

2

The products that are in the double summations can be featured in the following double array (for $n=4$):

$$\begin{array}{|l||l|l|l|l|} \hline &\color{red}{\binom{4}{0}}&\color{red}{\binom{4}{1}}&\color{red}{\binom{4}{2}}&\color{red}{\binom{4}{3}}&\color{red}{\binom{4}{4}}\\ \hline \color{red}{\binom{4}{0}}&&*&*&*&*\\ \hline \color{red}{\binom{4}{1}}&&&*&*&*\\ \hline \color{red}{\binom{4}{2}}&&&&*&*\\ \hline \color{red}{\binom{4}{3}}&&&&&*\\ \hline \color{red}{\binom{4}{4}}&&&&&\\ \hline \end{array}$$

The first sum uses all the boxes. The second sum, assumed to be

$$\sum \sum \binom{n}{r}\binom{n}{s} \ \text{for} \ 0 \leq r < s\leq n$$

uses only the starred boxes.

Jean Marie
  • 81,803
1

The following representations of the double sums might be useful.

We have \begin{align*} \sum_{r=0}^n\sum_{s=0}^n\binom{n}{r}\binom{n}{s}&=\sum_{\color{blue}{0\leq r,s\leq n}}\binom{n}{r}\binom{n}{s} =\sum_{s=0}^n\sum_{r=0}^n\binom{n}{r}\binom{n}{s}\\ \sum_{r=0}^{n-1}\sum_{s=r+1}^n\binom{n}{r}\binom{n}{s}&=\sum_{\color{blue}{0\leq r<s\leq n}}\binom{n}{r}\binom{n}{s} =\sum_{s=1}^n\sum_{r=0}^{s-1}\binom{n}{r}\binom{n}{s}\ \end{align*}

Markus Scheuer
  • 108,315
0

$$\sum_{r=0}^n\sum_{s=0}^n{n\choose r}{n\choose s}=2\sum_{r=0}^{n-1}\sum_{s=r+1}^n{n\choose r}{n\choose s}+\sum_{t=0}^n{n\choose t}^2$$

Qwerty
  • 6,165
  • I have assumed that $0\leq r\lt s\leq n$ just to give the idea. Otherwise change limits accordingly. – Qwerty Nov 30 '17 at 07:04