Find the value of $$\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j} $$

I get the result: $$\frac{1}{2}\left(2^{2n}-\binom{2n}{n}\right)$$ via a numeric argument.

My question is: Can we solve it using a combinational argument?

My Numeric Argument: $$\left(\sum^{n}_{r=0}\binom{n}{i}\right)^2=\sum^{n}_{r=0}\binom{n}{i}^2+2\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j}$$

So here $$\displaystyle \sum^{n}_{r=0}\binom{n}{i} = \binom{n}{0}+\binom{n}{1}+.....+\binom{n}{n} = 2^n$$

and $$\displaystyle \sum^{n}_{r=0}\binom{n}{i}^2=\binom{n}{0}^2+\binom{n}{1}^2+.....+\binom{n}{n}^2 = \binom{2n}{n}$$

above we have calculate Using $$(1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+.....+\binom{n}{n}x^n$$

and $$(x+1)^n = \binom{n}{0}x^n+\binom{n}{1}x^{n-1}+\binom{n}{2}x^{n-2}+.....+\binom{n}{n}x^0$$

Now calcualting Coefficient of $x^n$ in $$(1+x)^n\cdot (x+1)^n = (1+x)^{2n} = \binom{2n}{n}$$

So we get $$\mathop{\sum\sum}_{0\leq i<j\leq n}\binom{n}{i}\cdot \binom{n}{j} = \frac{1}{2}\left[2^{2n} - \binom{2n}{n}\right]$$


  • $\begingroup$ Really sloppy to write $(1+x)^{2n}=\binom{2n}{n}$. $\endgroup$ – Thomas Andrews Feb 27 '16 at 17:35
  • $\begingroup$ I don't agree on how you obtain $\binom{2n}{n}$. What did you do with the mixed terms? EDIT: Now, I see what you did. You only look at the coefficients for $x^n$. Nice. $\endgroup$ – Friedrich Philipp Feb 27 '16 at 17:36
  • $\begingroup$ Mixed terms? @FriedrichPhilipp $\endgroup$ – Thomas Andrews Feb 27 '16 at 17:36
  • $\begingroup$ Why don't you ask up front for a combinatorial proof, if that is your main question, rather than forcing answerers to read a non-combinatorial proof? $\endgroup$ – Thomas Andrews Feb 27 '16 at 17:39
  • $\begingroup$ @Thomas Andrews: Yes. In $(1+x)^n(1+x)^n$. But I edited my comment. $\endgroup$ – Friedrich Philipp Feb 27 '16 at 17:39

Consider two sets, $A$ and $B$ each with $n$ elements. All elements are considered distinct.

$\displaystyle \sum_{0 \leq i < j \leq n} \binom{n}{i} \binom{n}{j}$ can be interpreted as the number of ways to pick a non-empty subset of $A \cup B$ with the requirement that the number of elements from $A$ who are picked is strictly smaller than the number of elements from $B$ who are picked.

$2^{2n}$ counts the total number of ways to pick a subset of any size from $A \cup B$. The number of cases where the same number of elements are picked from $A$ and $B$ (including the empty set) is obtained from the sum $\displaystyle \sum_{i=0}^n \binom{n}{i}^2$.

By symmetry, half of the $\displaystyle 2^{2n} - \sum_{i=0}^n \binom{n}{i}^2$ cases have more elements from $A$ compared to $B$.

The identity $\displaystyle \sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}$ matches the result with yours.

I do not know of a combinatorial argument for this last identity though. Does anyone have any?

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    $\begingroup$ The last can be rewritten $\sum \binom{n}{i}\binom{n}{n-i}$ and the terms represent the number of ways to choose $i$ elements from $A$ and $n-i$ elements from $B$, which, when summed, is the number of ways to choose $n$ elements from $A\cup B$. $\endgroup$ – Thomas Andrews Feb 27 '16 at 17:44
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    $\begingroup$ How many ways to choose $n$ people from $n$ boys and $n$ girls? For every decision about which $i$ boys we will pick, there are $\binom{n}{i}$ ways to decide which girls we will not pick. $\endgroup$ – André Nicolas Feb 27 '16 at 17:48

A bijective correspondence can be established between this issue and the following one:

[Dealing with the LHS of the equation :]

Let $S$ be a set with Card(S)=n.

Consider all (ordered) pairs of subsets $(A,B)$ such that

$$A \subsetneqq B \subset S. \ \ (1)$$

[Dealing with the RHS of the equation :]

Consider all subsets of a set $T$ with $2n$ elements, then exclude a certain number of them (to be precised later), $T$ being defined as :

$$T:=S \cup I \ \ \ \ \text{with} \ \ \ \ \ I:=\{1,2,\cdots n\}.$$

Let $C$ be any subset of $T$. We are going to establish (in the "good cases") a correspondence between $C$ and an ordered pair $(A,B)$ verifying (1).

Let us define first a certain fixed ordering of the elements of $S$ :

$$a_1 < a_2 < \cdots < a_n. \ \ (2)$$

Let $B:=T \cap S$ and $J:=T \cap I$. Three cases occur :

  • If $Card(J)<Card(B)$, $J$ is the set of indices "selecting" the elements of $B$ that belong to $A$ in the ordered set $S$.

  • If $Card(J)>Card(B)$, we switch the rôles of indices and elements. This accounts for the half part of the formula: indeed this second operation will give the same sets $(A,B)$.

  • If $Card(J)=Card(B)$, which happens in $2n \choose n$ cases, such cases cannot be placed in correspondence with a case considered in (1), thus have to be discarded.

I know this could be written in a more rigorous way, but I believe the main explanations are there.

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