Intuition on the sum of first (n-1) numbers is equal to the number of ways of picking 2 items out of n. 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?
 A: 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...
A: 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!
A: Wolfram proof without words. Note that this uses Pascal's Triangle.
A: Let's count 2-subsets from a set of size $n$.
Certainly this is $n\choose2$.
Alternatively, we can count by cases.  First some details:
Let $N$ be the set $\{x_1, x_2, \ldots x_{n}\}$ where $x_i < x_{i+1}$ for $1 \leq i \leq n - 1$.  
We seek to count how many unique subsets $\{a, b\} \subseteq N$ there exist where $a < b$.
Cases:
1) $b = x_n$ $\hspace{14mm}$  As $a < b$, there remain $(n - 1)$ choices for a.
2) $b = x_{n-1}$ $\hspace{9mm}$  There remain $(n - 2)$ choices for $a$.
3) $b = x_{n-2}$ $\hspace{9mm}$  There remain $(n - 3)$ choices for $a$.
$\hspace{10mm}\vdots$
$n$) $b = x_0$ $\hspace{13mm}$  There remain $0$ choices for $a$.
Summing over each of these distinct cases gives us $\sum_{i=0}^{n-1}i$.
Hence, $\frac{n(n-1)}{2} =$ $n\choose2$ $= 1 + 2 + \ldots (n-1)$.
