# Binomial Coefficient with Two

## Theorem

$\forall r \in \R: \dbinom r 2 = \dfrac {r \left({r - 1}\right)} 2$

### Corollary

The usual presentation of this result is:

$\forall n \in \N: \dbinom n 2 = T_{n - 1} = \dfrac {n \paren {n - 1} } 2$

where $T_n$ is the $n$th triangular number.

## Proof

From the definition of binomial coefficients:

$\dbinom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$

where $r^{\underline k}$ is the falling factorial.

In turn:

$\displaystyle x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \left({x - j}\right)$

When $k = 2$:

$\displaystyle \prod_{j \mathop = 0}^1 \left({x - j}\right) = \frac {\left({x - 0}\right) \left({x - 1}\right)} {2!}$

where $2! = 1 \times 2 = 2$.

So:

$\forall r \in \R: \dbinom r 2 = \dfrac {r \left({r - 1}\right)} 2$

$\blacksquare$