TITLE: Exactly how does the equation n(n-1)/2 determine the number of pairs of a given number of items (n)?
QUESTION [15 upvotes]: I understand the purpose for division. Multiplication is simply repeated addition and division is simply repeated subtraction. So lets say we have 4 total items. Plugging 4 into the equation we get 4(4-1)/2 = 12/2 = 6.
So there are 6 possible combinations with 4 items. Applying the intuitive understanding of division as repeated subtraction, we can plot 12 on a numberline, and then since we are dividing by 2, we count backwards by 2 until we reach 0. We can do this 6 times. 
That makes sense to me. My issue is with the numerator. Why must a number of items be multiplied by one number less than itself? I understand the basic reason for multiplying the number of items by itself: pairs include two items, so the 4 takes care of the possible numbers on the first item, but why does the second item only receive 3 possible combinations? My hypothesis is that it prevents overlap, but I am having difficulty interpreting the logic of this. How exactly does it overlap? Can this be visualized on a matrix? Why does 4*4/2 = 8 not give the correct number of unique pairs?

REPLY [3 votes]: You are almost correct that $4*4/2$ would give us the amount of pairs if we simply chose two numbers from our set of n numbers, and let order matter.
Let $[a,b,c,d]$ be our set of four objects. According to your plan, we pick two of the objects from this set with replacement, meaning that if we choose $a$, then we can choose $a$ again as our second object. This would give us the following pairs.
$$ aa,ab,ac,ad\\
ba,bb,bc,bd\\
ca,cb,cc,cd\\
da,db,dc,dd $$
As you can see, we do count each of the numbers that are not in the main diagonal twice, but the numbers on the diagonal are counted only once. Our square here is $4$ by $4$, so there are $4$ numbers on the diagonal and $16-4=12$ numbers that are not. So we divide the second number by two and add it to the first to get the total amount of distinct pairs without order mattering, would in fact be $10$, and for a general $n$ by $n$ square of possibilities, you can see that this number becomes $n(n-1)/2+n$.
However, I think the problem that you are being asked here is how many ways are there to choose two numbers from the set with replacement. Thus we would simply subtract the diagonal from our previous answer to get $n(n-1)/2$.
Hope this helps!