Proof with Combinatorial Argument $\sum_{i = 1}^{n} (i-1) = nC2$

Question

I am trying to prove below equation with combinatorial argument but I have no idea how this works.

$$\sum_{i = 1}^{n} (i-1) = nC2$$

Can anyone give me a clue?

Answer 1

Let's think of the following problem. Given n points ($p_1,p_2,\ldots , p_n$) how many lines can be formed that join exactly 2 of those n points?. Well from $p_1$ we can draw $n-1$ lines, one joining it to $p_2$, another to $p_3$ and so on. Then from $p_2$ we can draw another $n-2$ lines, joining it to $p_3,p_4\ldots$.

So the total number of lines is $\sum\limits_{i=1}^{n}{(i-1)}$.

Can you think of a combinatorial argument for the total number of lines for this problem ?

Answer 2

Hint. Two ways of counting handshakes between $n$ persons.

• Take the number of handshakes between $2$ out of $n$ people: $\binom{n}{2}$
• The first person shakes hands of $n-1$, the second person shakes hands of $n-2$, and so on.