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!
 A: 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 $(N+1)(N+2)/2$. 
A: 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 $${3+L-3m-3-1\choose L-3m-3}={L-3m-1\choose L-3m-3}=\frac{(L-3m-1)!}{2!(L-3m-3)!}={L-3m-1\choose 2}$$
A: *

*Replace $L$ by $L-3m$ and $x_i$ by $x_i-m$, you can assume that $m=0$.

*Let $$
f(L, p) = card\{ x\in (\Bbb N-\{0\})^p:\sum x_k = L
\}
$$You easily have:
$$
f(L,2) = L-1\\
f(L,p+1) = \sum_{x=1}^{L-p} f(L-x,p)
$$
In your case:


$$
f(L,3)= \sum_{x=1}^{L-2} f(L-x,2)\\
= \sum_{x=1}^{L-2} [L-x-1]\\
= \sum_{y=1}^{L-2} y = \frac 12(L-1)(L-2)
$$


*

*Going back to the initial problem, you get
$$
\frac 12(L-3m-1)(L-3m-2)
$$

A: You could use a generating function to solve this. Consider an individual $x_{i}$ term written as $f(x) = \sum_{j=m+1}^{\infty} x^{j}$. We then factor out $x^{m+1}$ to get $f(x) = x^{m+1} \sum_{i=0}^{\infty} x^{i}$. So our generating function $g(x) = x^{3m+3} (\frac{1}{1-x})^{3}$. Thus, it suffices to expand out $(\frac{1}{1-x})^{3}$ using the binomial identity: 
$$\frac{1}{(1-x)^{r}} = \sum_{i=0}^{\infty} \binom{i + r - 1}{i} x^{i}$$. 
Considering the coefficient of $x^{L - 3m - 3}$ gives us $\binom{L - 3m - 1}{L - 3m - 3}$.
