How find Number of integer solutions $x_{1}+x_{2}+\cdots+x_{n}=m$ Question:
let $m$ and $n$ be positive integers,The number of positive integer solutions to the equation

$$x_{1}+x_{2}+\cdots+x_{n}=m,m\ge n,x_{i}\ge 1,1\le x_{1}\le x_{2}\le\cdots\le x_{n},(i=1,2,\cdots,n)$$
  is $f(m,n)$,How find this close form $f(m,n)?$

I know this therom:

Let $n$ and $m$ be positive integers,The number of positive integer solution to the equation
  $$x_{1}+x_{2}+\cdots+x_{m}=n(n\ge m)$$ is
  $$N=\binom{n-1}{m-1}$$

But for my problem,I can't prove it.Thank you 
BY the way
In 2010 China math competition,there is this problem

find the total  number of sets of positive integer $(x,y,z)$,where $x,y$ and $z$ are positive integers, with  $x\le y\le z$
  and such that
  $$x+y+z=2010$$
this answer is $$336675$$

can see this PDF：(problem 8) http://wenku.baidu.com/view/9a59934ee518964bcf847cba.html
and I also Find this same problem is Sinapore Mathematical Olympiad (SMO)2012 Problem 4:

find the total  number of sets of positive integer $(x,y,z)$,where $x,y$ and $z$ are positive integers, with  $x\le y\le z$
  and such that
  $$x+y+z=203$$

This follow is office Solution(Maybe is wrong).
First note that there are $\binom{202}{2}=\dfrac{202(201)}{2}=20301$ positive integers sets$(x,y,z)$ which satisfy the given equation.These solution sets include those where two of the three values are equal.if $x=y$ then $2x+z=203$,By enumeerating,$z=1,3,5,\cdots,201$.There are thus $101$ solutions of the form $(x,x,z)$, similarly,there are $101$ solutions of the form $(x,y,x)$ and $(x,y,y)$,since $x<y<z$,the required answer is
$$\dfrac{1}{3!}\left(\binom{202}{2}-3(101)\right)=\dfrac{20301-303}{6}=3333$$
 A: You are looking for the partition of $m$ in $n$ positive parts.
You can solve this using a generating function, taking the coefficient of $x^m$ in the expansion of $$\frac{x^n}{\displaystyle\prod_{i=1}^{n} (1-x^i)}$$ or by recursion, which is used in my Java applet at http://www.se16.info/js/partitions.htm : so if you ask for partitions of $2010$ into exactly $3$ parts you get a result of $336675$; partitions of $203$ into exactly $3$ parts gives $3434$.
Looking at what you call the "office solution", there are indeed $20301$ compositions of $203$ into $3$ positive parts and $3 \times 101$ of these have two parts the same.  So there are $\frac{20301-3\times 101}{3!} = 3333$ partitions of $203$ into $3$ distinct positive parts.  But there are also $101$ partitions of $203$ into $3$ positive parts where two (but not three) are identical, giving the correct answer of $3333+101=3434$.    
A: The theorem gives you the number of solutions of $x_{1} + \cdots + x_{n} = m$ that are permutations dependent ; this is called the stars & bars theorem. For example, $(1;2;1;2)$ and $(1;1;2;2)$ are 2 distinct solutions if $n=4$ and $m=6$.
However here you would only consider increasing sequences as solutions. These are called integer partitions. More precisely you want to find the number $f(m,n)$ of partitions of $m$ in $n$ parts $\geq 1$. Unfortunately, as far as I know, there is no (simple) closed form of $f(m,n)$. Its generating function, on the other hand, has a simple expression
$$\sum_{m,n=1}^{\infty} f(m,n) x^{n}q^{m} = \prod_{m=1}^{\infty} \frac{1}{1-xq^{m}}$$
To find $f(m,n)$, one can then compute the term in front of $x^{n}q^{m}$ in the RHS. One can then use Taylor expansion to find it numerically.
Some closed forms of $f(m,n)$
The wikipedia article about integer partitions happens to have closed forms of $f(m,n)$ for the first few values of $n$

*

*$n=1$ : Obviously for all $m$, $f(m,1) =1$

*$n=2$ :
$$f(m,2)=\left\lfloor \frac{m}{2} \right\rfloor$$

*$n\leq 3$ : According to Hardy's paper Some Famous Problems of the Theory of Numbers, $\displaystyle f(m,n\leq 3) = \sum_{n=1}^{3}f(m,n)$ is the nearest integer to
$$\frac{(m+3)^2}{12}$$
