Proving Pascal's Rule : ${{n} \choose {r}}={{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$ I'm trying to prove that ${n \choose r}$ is equal to ${{n-1} \choose {r-1}}+{{n-1} \choose r}$ when $1\leq r\leq n$.
I suppose I could use the counting rules in probability, perhaps combination= ${{n} \choose {r}}=\frac{n!}{r!(n-r!)}$.
I want to see an actual proof behind this equation.  Does anyone have any ideas?

How about this:
Note that
$${n \choose r} = \frac {n(n-1)(n-2)...(n-r+1)} {1\cdot2\cdot3\cdot...(r-1)r}$$
This can be written as 
$${n \choose r} = \frac nr \cdot \frac {(n-1)(n-2)...(n-r+1)} {1\cdot2\cdot3\cdot...(r-1)} = \frac nr {n-1 \choose r-1} $$
and also 
$${n \choose r} = \frac n{n-r} \cdot \frac {(n-1)(n-2)...(n-r+1)(n-r)} {1\cdot2\cdot3\cdot...(r-1)r}= \frac n{n-r} {n-1 \choose r}$$
So
$$\begin {align}
{n-1 \choose r-1} + {n-1 \choose r} & = \frac rn {n\choose r} + \frac {n-r} n {n\choose r} \\
& = {n \choose r}
\end {align} $$
as required.
