Calculating summation involving binomial coefficients

Question

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

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

Answer 1

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
$\displaystyle \frac{12,952}{2} = 6,476.$

Answer 2

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 $0\leq i\leq 6$.

Answer 3

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