Intuition on the sum of first (n-1) numbers is equal to the number of ways of picking 2 items out of n.

Question

While going through an equation today i realized that sum of first (n-1) numbers is [n*(n-1)/2] which is equal to combinations of two items out of n i.e [n!/((n-2)! * 2!)]. I need some intuition on how these two things are related?

Answer 1

Wolfram proof without words. Note that this uses Pascal's Triangle.

Answer 2

One way to think of this is to map it to an intermediary representation - namely, a triangle made of boxes:

 *****
 ****
 ***
 **
 *

Let's suppose that this triangle has width and height $n$. The number of dots in this triangle is equal to $1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$.

We can interpret this triangle as every way of choosing two elements out of $n + 1$ by expanding it out into a 0/1 matrix:

    | 1   2  ... n-1  n
----+------------------
n+1 | 1   1   1   1   1
n   | 1   1   1   1   0
n-1 | 1   1   1   0   0
... | 1   1   0   0   0
2   | 1   0   0   0   0
1   | 0   0   0   0   0

If we take all unordered pairs of two numbers, then we can always sort the pair by putting the bigger number first. Each possible way to do this corresponds to a 1 entry in this matrix. For example, the pairing $(n+1, n)$ corresponds to the upper-right corner, since $n+1 > n$. Similarly, $(n+1, 1)$ corresponds to the top-left corner. If you count up the number of 1s in this matrix, you'll note that it's half the area of the matrix. There are $n + 1$ rows and $n$ columns, so the area is $\frac{n(n + 1)}{2}$.

We can also arrive here by noting that there are $n$ 1s in the first row, then $n - 1$, then $n - 2$, ..., then 1. Thus $1 + 2 + ... + n$ is equal to the number of unordered pairs drawn from $n + 1$ numbers.

Hope this helps!

Answer 3

You're asking why the number of ways to pick 2 cards out of a deck of n is the same as the sum 1 + 2 + ... + (n-1).

The reason is that there are (n-1) ways to pair the first card with another card, plus (n-2) ways to pair the second card with one of the remaining cards, plus (n-3) ways to pair the third card...