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$$\sum_{0<=i<j<=7} \sum_{}\binom{7}{i} \binom{7}{j}$$

I'm not aware how to we proceed with such summations what do they represent so any way to teach me out this summation

imposter
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    The computation becomes easier if you use the $(7)$-th row of Pascal's Triangle (i.e. ${1,7,21, \cdots}$). Then, if it were me, I would ignore any hidden meaning of the computation, avoid any attempt at elegance, and simply manually compute the double summation that will end up having $(8 + 7 + \cdots + 2 + 1) = 28$ terms. Each term will be the product of two numbers, where each number is taken from one of the elements in the 7th row of Pascal's triangle. – user2661923 Apr 27 '21 at 01:20
  • Re previous comment, you can use the distributive property as a minor shortcut. That is (for example) $(a \times b) + (a \times c) + (a \times d) + (a \times e) = [a \times (b + c + d + e)].$ – user2661923 Apr 27 '21 at 01:22
  • By the way, re my first comment, I misinterpreted the summation as $\cdots i \leq j\cdots$ rather than $\cdots i < j \cdots$. Therefore, you only have $(7 + 6 + \cdots + 2 + 1) = 21$ terms. – user2661923 Apr 27 '21 at 01:29
  • Well, this is embarrassing. It seems that $(7 + 6 + \cdots + 2 +1) = 28.$ Go figure. – user2661923 Apr 27 '21 at 02:54

3 Answers3

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Actually, I just thought of a shortcut.

Consider $(1 + 7 + 21 + 35 + 35 + 21 + 7 + 1)\times (1 + 7 + 21 + 35 + 35 + 21 + 7 + 1).$

This product, which has $(64)$ terms, is equal to $(2^7)^2 = (128)^2 = 16,384.$

This represents $i \in$ {$0,1,\cdots,7$} and $j \in$ {$0,1,\cdots,7$}.

Deduct from this the $8$ times that $i = j$. This is represented by

$(1^2 + 7^2 + 21^2 + 35^2 + 35^2 + 21^2 + 7^2 + 1^2)$

$= 2 \times (1^2 + 7^2 + 21^2 + 35^2) = 2 \times 1,716 = 3,432.$

$16,384 - 3,432 = 12,952$.

This represents the $(56)$ terms where
$i \in$ {$0,1,\cdots,7$} and $j \in$ {$0,1,\cdots,7$},
and $i \neq j$.

This must be exactly double the sum, since by symmetry, each occurrence of $i < j$ is balanced by an occurrence of $i > j$.

Therefore the desired computation of the 28 terms is
$\displaystyle \frac{12,952}{2} = 6,476.$

user2661923
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Here’s a faster way to calculate user2661923’s solution.

The sum of 7 choose i from 0 to 7 is $(1+1)^7=2^7=128$.

The sum of 7 choose i squared is 14 choose 7=3432 (i from first 7, i not to choose from second 7).

This gives $(128^2-3432)/2=6476$

Eric
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Summations like this are funny-looking. I don't think we need two sigmas here. $$\sum_{0\leq i<j\leq 7} \binom{7}{i} \binom{7}{j}.$$

I imagine we are talking about $i,j$ integers. First look inside the summation:

$$\binom{7}{i} \binom{7}{j}.$$

We will sum a certain collection of expressions like this. To evaluate just one of these, we need to know the values of integers $i$ and $j$.

Which pairs of integers $i,j$ will occur in the sum?

Those, and only those, satisfying $0\leq i<j\leq 7$.

This can be expressed as a double sum, similar to double integrals in calculus.

To separate into a double sum, you could note that $0\leq i\leq 6$.

311411
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  • I prefer two sigmas. Neat explanation tho – Vons Apr 27 '21 at 01:30
  • The two-sigma notation doesn't bother me. But in my mind, this is just an instance of $\sum_{t,\in,\Omega} a(t),$ where $\Omega$ is a set and $a$ is a given function. – 311411 Apr 27 '21 at 01:35