Proving a Combinatorial Theorem The Theorem

My Problem
I don't really understand how the $RHS$ counts the number of final positions for a $1$. I understand how summing all of these cases would be the same as counting all the actual strings, but not how the $RHS$ counts the number of final positions for a $1$.
 A: Suppose $r=3$,  $n=9$. We want to count the number of bit strings of length 10 containing 4 ones. Now, the last one can be in position 4 to position 10 ($k=r+1 \cdots n+1$). Por example, how many strings are with the last one in position 6 $(k=6)$? We have
  * * * * * 1 0 0 0 0 

where the 5 ($k-1$) asterisks must contain 3 ($r$) ones. So we have ${k-1 \choose r}={6 \choose 3}$ alternatives. We must sum that from $k=r+1$ to $k= n+1$
A: For any bit string of length n+1 containing r+1 ones, where r+1 $\le$ n+1, the final position for a one will range from position r+1 to position n+1 (inclusively). If the final one is situated at position n+1, then there are C(n, r) such bit strings of length n+1 containing r+1 ones (because we need to choose the remaining r ones from the remaining n positions). Now, if if the final one is situated at position n, then there are C(n-1, r) such bit strings of length n+1 containing r+1 ones. If we keep this up, we are essentially applying a summation of C(j, r) from j = r to j = n. The result gives us the total number of bit strings of length n+1 containing r+1 ones.
