Equivalence of another formula for the number of $r$-combinations with repetition allowed 
Basically it means choosing r things out of n, where order doesn't matter, and you are allowed to pick a thing more than once. For example, $\{1, 1, 2\}$ out of $\{1, 2, 3, 4\}$.
I managed to find another solution:
$$
{n \choose r} + (r-1){n \choose r-1} + (r-2){n \choose r-2} + \cdots + {n \choose 1}
$$
I am having trouble proving that these two are equivalent. 

As has already been pointed out, unfortunately  the two solutions are not equivalent.
However, if we use Pascal's Rule:-
$${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$
and apply this $r$ times to 
${r+n-1 \choose r}$ the following solution can be shown to be equivalent:- 
$${r-1 \choose r-1}{n \choose r} + {r-1 \choose r-2}{n \choose r-1} + {r-1 \choose r-3}{n \choose r-2} + \cdots + {r-1 \choose 0}{n \choose 1}$$
In other words the following relationship holds:-
$${r+n-1 \choose r}=\sum_{k=1}^r{r-1 \choose k-1}{n \choose k}$$
Perhaps allowing the repetition of multiple elements at the same time results in the binomial terms ${r-1 \choose k-1}$ for $k\in \{1,2,..,r\}.$
