Vtyjkyuk

I can see this easily by definition. But will it have combinatorial meaning of $binomn2 = sum_k=1 ^n-1 (k)?$

There is a 1-to-1 correspondence between each blue pair, and each yellow circle. With $n$ blue circles there are $binomn2$ blue pairs, and the number of yellow circles is $sum_i=1^n-1i$.

Clearly $sumlimits_i=0^n-1 i=sumlimits_j=1^n (j-1)$ by using the substitution $j=1+1$

A combinatorial demonstration of $n choose 2 = sumlimits_j=1^n (j-1)$:

• $n choose 2$ is the number of ways of choosing two integers from $1,2,ldots,n$


• if the larger of the two integers is $j in 1,2,ldots,n$ then there are $j-1$ possibilities for the smaller of the two integers, so in total there are $sumlimits_j=1^n (j-1)$ possible pairs


• so these two expressions are equal

A geometric illustration this is $fracn(n-1)2$:

Consider $dbinomn+12$. It is the number of ways to choose two items from a set of $n+1$ items.

So, start with a single item (call it item 1). There are $n$ other items that can be picked to pair with it.

Next, suppose we do NOT pick item 1. We start with item 2. Now, there are $n-1$ other items (that are not item 1 nor item 2) that can be picked to pair with it.

...

Suppose we do NOT pick items 1 through $n-1$. We pick item $n$. There is exactly 1 other item (not items 1 through $n$) that can be picked to pair with it.

$sum_i=1^n$ is $binomn+12$, not $binomn2$.

Consider the set of $n=1$ tokens $1,2,ldots,n-1,n,*$. $binomn+12$ is the size of the set of possible pairs $(i,j)$ with $i<j$ (and $*$ greater than all the rest). Each pair $(i,j)$ represents a point on the plane, and each pair $(i,*)$ represents the point $(i,i)$. These points visualize as:

$$beginarrayccccc
star&star&cdots&star&star\
star&star&cdots&star\
vdots&vdots\
star&star\
star
endarray$$

Counting from the bottom up, you have $1+2+cdots+(n-1)+n$ points. So $binomn+12=sum_i=1^ni$.