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!

It depends on the number $L,m$. If $L=1,m=2$ then your set is empty. Totally if $L<m$ then your set is empty.
Even we must have $L=x_1+x_2+x_3\geq (m+1)+(m+1)+(m+1)=3m+3$ i.e $L\geq 3m+3$.
Hence I change your question to this that:
What is the cardinality of the set $X=:\{(x_1,x_2,x_3)\in \mathbb{N}^3:\ x_1+x_2+x_3=L,\ x_i>m\}$ when $L,m\in\mathbb{N}$ with  $L\geq 3m+3$.
And my answer is this that cardinality of $X$ is equal to cardinality of the set $\tilde{X}=:\{(x_1,x_2,x_3)\in (\mathbb{N}\cup\{0\})^3:\ x_1+x_2+x_3=L-3m-3\}$. For example when $L=9,m=2$ then the cardinality you want is the cardinality of the set $\tilde{X}\{(x_1,x_2,x_3)\in (\mathbb{N}\cup\{0\})^3:\ x_1+x_2+x_3=0\}$which is one and it's element is $(0,0,0)$ and it's related element in $X=\{(x_1,x_2,x_3)\in \mathbb{N}^3:\ x_1+x_2+x_3=9,\ x_i>2\}$ is $(3,3,3)$. Or  if $L=10,m=2$ the set $\{(x_1,x_2,x_3)\in (\mathbb{N}\cup\{0\})^3:\ x_1+x_2+x_3=1\}$'s cardinality is 3 and it's elements are $(1,0,0),(0,1,0),(0,0,1)$ and their related elements in $X=\{(x_1,x_2,x_3)\in \mathbb{N}^3:\ x_1+x_2+x_3=10,\ x_i>2\}$ is $(4,3,3),(3,4,3),(3,3,4)$.
But there exists a formula in discrete mathmatics which gives us the cardinality of $\tilde{X}=:\{(x_1,x_2,x_3)\in (\mathbb{N}\cup\{0\})^3:\ x_1+x_2+x_3=L-3m-3\}$ when $L-3m-3\geq0$. For this special problem answer is $${3+L-3m-3-1\choose L-3m-3}={L-3m-1\choose L-3m-3}=\frac{(L-3m-1)!}{2!(L-3m-3)!}={L-3m-1\choose 2}$$
