Explain a couple steps in proof that ${n \choose r} = {n-1 \choose r-1} + {n-1 \choose r}$ Show  ${n \choose r}$ = ${n-1 \choose r-1}$ + ${n-1 \choose r}$ 
I found a similar question on here but I am looking for a little bit more of an explanation on how they simplified 
Right Hand Side $$= \frac{(n-1)!}{(r-1)!(n-r)!} + \frac{(n-1)!}{r!(n-r-1)!}$$
                $$= \frac{r(n-1)!+(n-r)(n-1)!}{r!(n-r)!}$$
               $$= \frac{n!}{r!(n-r)!}$$
               $$= \binom{n}{r}$$,
the first equality because of definition
,the second equality because of summing fractions
the third because of $n(n-1)!=n!$,
the fourth by definition.
I would like more of an explanation on equality 2 and equality 3.  I understand that they are summing fractions but I am interested in how they came up with the common denominator and then simplified. 
 A: The common denominator comes from observing $(r-1)! \times r = r!$ and $(n-r-1)! \times (n-r)=(n-r)!$.
The sum comes from observing $r(n-1)!+(n-r)(n-1)!=(r+n-r)(n-1)!=n (n-1)!=n!$.
A: Hint: You can just think of this as you can mark one of your elements, in the first case, you choose it, in the second case you don't choose it, the sum on the right side is equal to that case.
If you choose it, then you choose $r-1$ elements from $n-1$ elements.
If you don't choose it, then you choose $r$ elements from $n-1$ elements.
Adding up these two cases equal the left side.
A: Here's a more detailed breakdown of the summation. I added a few steps to show finding the common denominator and adding up/factoring the two numerators.
$$
\begin{align}
\dbinom{n-1}{r-1} + \dbinom{n-1}{r} &= \dfrac{(n-1)!}{(r-1)!(n-r)!} +  \dfrac{(n-1)!}{r!(n-r - 1)!} \\
&= \dfrac{r}{r} \cdot \dfrac{(n-1)!}{(r-1)!(n-r)!} + \dfrac{n - r}{n - r} \cdot \dfrac{(n-1)!}{r!(n-r - 1)!} \\
&= \dfrac{r(n-1)!}{r!(n-r)!} +  \dfrac{(n-r)(n-1)!}{r!(n-r)!} \\
&= \dfrac{r(n-1)! + (n-r)(n-1)!}{r!(n-r)!} \\
&= \dfrac{(n-1)![r + (n-r)]}{r!(n-r)!} \\
&= \dfrac{(n-1)! \cdot n}{r!(n-r)!} \\
&= \dfrac{n!}{r!(n-r)!} \\
&= \dbinom{n}{r}
\end{align}
$$
A: For a combinatorial proof, $\binom nr$ is the number of $r$-element subsets of $\{1,2,\ldots,n\}$. Each such subset either contains $n$ or does not. $\binom {n-1}{r-1}$ counts the former, $\binom {n-1}r$ counts the latter. Therefore the quantities $\binom nr$ and $\binom {n-1}{r-1} + \binom {n-1}r$ are equal.
