Number of solutions of $ x_1 + x_2 + ... + x_g \le n$ I'm trying to prove that the number of solutions of 
$$ x_1 + x_2 + ... + x_g \le n$$
is 
$$ \dbinom{g + n}{n} $$
so far I've been able to show that the number of solutions of
$$ x_1 + x_2 + ... + x_g = n$$
is 
$$ \dbinom{g + n - 1}{n} $$
But I can't manage to see how to get from my result to the goal result. I used a partitioning argument for my result. Is this the wrong way of going about it?
Thanks!

There are two ways of doing this. The first way is to add them up as Ross Millikan suggested, to get
$$\binom{g-1}{0}+\binom{g}{1}+\binom{g+1}{2}+\cdots+\binom{g+n-1}{n}$$
as your answer. The second way is to see that there is a bijection between the set of solutions of $$x_1+\cdots+x_g \le n$$ and the set of solutions of $$x_1+\cdots+x_{g+1}=n.$$ Hint: If $(x_1,\dots,x_g)$ is satisfies the first inequality then $$(x_1,\dots,x_g,n-x_1-\cdots-x_g)$$ satisfies the second equation.
This in fact gives a proof of the identity
$$\sum_{k=0}^n \binom{g+k-1}{k} = \binom{g+n}{n}.$$
