Number of triplets adding to a certain number Suppose I have $L$ and $m$ in $\mathbb{N}$.
What is the cardinality of the set
$$
\{ (x_1, x_2, x_3) \in \mathbb{N}^3 : x_1 + x_2 + x_3 = L, x_i > m   \}?
$$
An exact number would be great but I would be quite happy with a lower bound also.
This is part of my homework, not this exact question. I am thinking about 
using this bound to solve my homework problem. Thanks!

If $L\lt 3(m+1)$, there are no solutions.
If $L\ge 3(m+1)$, let $M=L-3m$, and let $x_i=y_i+m$. We want to find the number of solutions of $y_1+y_2+y_3=M$ in positive integers.
Imagine that we write down $M$ copies of $\ast$, separated by some space, like this:
$$\ast\qquad \ast\qquad \ast\qquad \ast\qquad\ast\qquad \ast\qquad \ast\qquad \ast\qquad\ast\qquad \ast\qquad \ast\qquad \ast$$
These $M$ "stars" determine $M-1$ interstellar gaps. Choose $2$ of these gaps to insert separators into. These are traditionally called bars.
Every placement of the bars determines a positive integer solution of $y_1+y_2+y_3=M$. Just let $y_1$ be the number of stars until the first bar, $y_2$ the number of stars between the $2$ bars, and $y_3$ the number of stars after the last bar. Conversely, any solution in positive integers of $y_1+y_2+y_3$ determines a placement of bars.
Thus there are $\binom{M-1}{2}$ ways to do the job. 
Remark: For more information, please see the Wikipedia article on Stars and Bars.
Here we were dealing with the a sum $x_1+x_2+x_3$ of $3$ numbers. For this case, there is a simpler way to look at the problem. It is convenient to let $N=L-3(m+1)$. We want to find the number of solutions of $z_1+z_2+z_3=N$ in non-negative integers. Just count separately the solutions that have $z_1=0,1,2,\dots,N$ and add up. 
If $z_1=0$, then $z_2$ can have $N+1$ values, anything from $0$ to $N$
If $z_1=1$, then $z_2$ can have $N$ values.
And so on, until if $z_1=N$ then $z_2$ can have $1$ value. Thus the total is $(N+1)+N+(N-1)+\cdots+1$. This is (backwards) a familiar sum, which simplifies to $(N+1)(N+2)/2$. 
