TITLE: Number of total possibilities for an equation
QUESTION [10 upvotes]: I need to find the number of possibilities for which the following equation exists:
$$x_1 + x_2 + x_3 + \cdots + x_{10} \leq 70$$
Each variable is a non-negative integer.
I tried simplifying the question to the point of finding the number of possible solutions to each equation seperatly:
$$x_1 + x_2 + x_3 + \cdots + x_{10} = 70,$$ which is ${79}\choose{10}$.
$$x_1 + x_2 + x_3 \cdots + x_{10} = 69,$$ which is ${78}\choose{10}$.
$$x_1 + x_2 + x_3 + \cdots + x_{10} = 68,$$ which is ${77}\choose{10}$, and so forth.
Then I thought about adding all of these together. This seems a bit too much work considering that the textbook solution is: ${80}\choose{10}$, and I can't seem to figure out the idiomatic way of approaching these kind of questions. 
Any help is greatly appreciated.

REPLY [3 votes]: Here is another way using your original method. The number of solutions to 
$$
x_1+x_2+\dotsb+x_{10}=r\quad (0\leq r\leq 70)
$$
is
$$
\binom{r+10-1}{10-1}=\binom{r+9}{9}.
$$
Thus the number of solutions to
$$x_1 + x_2 + x_3 + \cdots + x_{10} \leq 70$$
is
$$
\sum_{r=0}^{70}\binom{r+9}{9}=
\sum_{k=9}^{79}\binom{k}{9}
=\sum_{k=9}^{79}\left[\binom{k+1}{10}-\binom{k}{10}\right]
=\binom{80}{10}
$$
as the sum telescopes.