MHB Evaluating a sum involving binomial coefficients

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The discussion focuses on evaluating the double sum involving binomial coefficients, specifically $$\mathop{\sum \sum}_{0\leq i<j\leq n} (-1)^{i-j+1}{n\choose i}{n\choose j}$$. Participants explore rewriting the sum and recognize that the summand is symmetric in indices i and j. By adding the sums for both configurations and including diagonal terms, they derive a relationship leading to the conclusion that the sum is equal to half the sum of the squares of binomial coefficients. Ultimately, the evaluation results in $$S = \frac{(2n)!}{2(n!)^2}$$, confirming the result through combinatorial identities.
Saitama
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Problem:
Evaluate
$$\mathop{\sum \sum}_{0\leq i<j\leq n} (-1)^{i-j+1}{n\choose i}{n\choose j}$$

Attempt:
I wrote the sum as:
$$\sum_{j=1}^{n} \sum_{i=0}^{j-1} (-1)^{i-j+1}{n\choose i}{n\choose j}$$
I am not sure how to proceed from here. I tried writing down a few terms but that doesn't seem to help.

Any help is appreciated. Thanks!
 
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To me, the innermost sum looks like a binomial expansion...
 
Prove It said:
To me, the innermost sum looks like a binomial expansion...

I don't think so...(Wondering)

The sum is
$$-\sum_{j=1}^{n} (-1)^i{n\choose i} \sum_{i=0}^{j-1} (-1)^j {n\choose j}$$
The limits for innermost sum varies from 0 to j-1, not 0 to n.
 
Pranav said:
Problem:
Evaluate
$$\mathop{\sum \sum}_{0\leq i<j\leq n} (-1)^{i-j+1}{n\choose i}{n\choose j}$$
First, $(-1)^{-j} = (-1)^{\,j}$, so we can write the sum as $$S = \mathop{\sum \sum}_{0\leqslant i<j\leqslant n} (-1)^{i+j+1}{n\choose i}{n\choose j}$$. The summand is then symmetric in $i$ and $j$, so the sum $S$ is equal to $$ \mathop{\sum \sum}_{0\leqslant j<i\leqslant n} (-1)^{i+j+1}{n\choose i}{n\choose j}$$ (with $i$ and $j$ interchanged). If we now add those two sums, and also throw in the diagonal terms (those with $i=j$ in the summand, namely $$ \sum_{i=0}^n -{n\choose i}^2$$), then we would get the sum over the whole $n\times n$ array of indices $\{(i,j): 1\leqslant i \leqslant n,\; 1 \leqslant j \leqslant n\}$. But that sum is $$-\sum_{i=0}^n (-1)^i{n\choose i} \sum_{j=0}^n (-1)^j{n\choose j}$$. And $$\sum_{i=0}^n (-1)^i{n\choose i} = 0$$, because it is the binomial expansion of $(1-1)^n$.

It follows that $$2S - \sum_{i=0}^n {n\choose i}^2 = 0$$, and so $$S = \frac12\sum_{i=0}^n {n\choose i}^2 = \frac{(2n)!}{2(n!)^2}$$ (see formula (9) here for the sum of the squares of binomial coefficients $n\choose i$).
 
Opalg said:
First, $(-1)^{-j} = (-1)^{\,j}$, so we can write the sum as $$S = \mathop{\sum \sum}_{0\leqslant i<j\leqslant n} (-1)^{i+j+1}{n\choose i}{n\choose j}$$. The summand is then symmetric in $i$ and $j$, so the sum $S$ is equal to $$ \mathop{\sum \sum}_{0\leqslant j<i\leqslant n} (-1)^{i+j+1}{n\choose i}{n\choose j}$$ (with $i$ and $j$ interchanged). If we now add those two sums, and also throw in the diagonal terms (those with $i=j$ in the summand, namely $$ \sum_{i=0}^n -{n\choose i}^2$$), then we would get the sum over the whole $n\times n$ array of indices $\{(i,j): 1\leqslant i \leqslant n,\; 1 \leqslant j \leqslant n\}$. But that sum is $$-\sum_{i=0}^n (-1)^i{n\choose i} \sum_{j=0}^n (-1)^j{n\choose j}$$. And $$\sum_{i=0}^n (-1)^i{n\choose i} = 0$$, because it is the binomial expansion of $(1-1)^n$.

It follows that $$2S - \sum_{i=0}^n {n\choose i}^2 = 0$$, and so $$S = \frac12\sum_{i=0}^n {n\choose i}^2 = \frac{(2n)!}{2(n!)^2}$$ (see formula (9) here for the sum of the squares of binomial coefficients $n\choose i$).

This is great, thank you Opalg! :) (Bow)
 
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