Number of integer solutions if $x_1\leq x_2\leq x_3\leq \cdots\leq x_r\leq k$ How would I count the number of integer solutions to:
$x_1+x_2+\cdots+x_r=k$, given fixed $r$ and $k$ and $0\leq x_1\leq x_2\leq \cdots \leq x_r\leq k$?
Attempt at solution (incorrect, see per comments): I was thinking maybe I could exchange this for the number of integer solutions to:
$y_1+y_2+\cdots+y_r=x_r$ where $y_1=x_1, y_{i>1}=x_i-x_{i-1}$, but then it gets a little messy because there are multiple possible values for $x_r$ for a given $k$... I think $x_r$ can be at minimum $k/r$ rounded down (edit: nope... rounded up?), in which case the answer would be $\sum_{[k/r]\leq j\leq k}\binom{j+r-1}{r-1}$ but this isn't the nicest expression...
Thanks for your help!
 A: You can count these partitions by recursion.  
Let $p_n(x,y)$ be the number of partitions of $x$ into up to $y$ parts where the largest part is less than or equal to $n$.  You can start with $p_0(x,y)=0$ when $x>0$, $p_n(x,y)=0$ when $x \lt 0$; and $p_n(0,0)=1$; and . The key element is $$p_n(x,y)=p_{n-1}(x,y)+p_{n}(x-n,y-1)$$
and you want $p_k(k,r)$.
I have a Java applet here which does the calculation: type $k$ next to "Partitions of:", change "Any number of terms" to "Maximum number of terms:" and type $r$, and finally click on "Calculate partitions".   
A: Let $p(k,r)$ be the number you are looking for.  
Your formula can't be right, because $p(k,r)$ is constant when $r\geq k$.  That is:
$$\forall r\geq k: p(k,r) = p(k,k)$$
That's the case because at most $k$ of the $x_i$ can be positive, so the only solutions when $r\geq k$ can be solutions when $r=k$ plus an additional $r-k$ zeros.
On the other hand, your formula increases for all $r$.
The generating function for $p(k,r)$ is:
$$\sum_{k,r} {p(k,r)x^ky^r} = \prod_{j=0}^\infty{\frac{1}{1-x^jy}}$$
It is unlikely you'll find a nice formula for $p(k,r)$, given their relation to the rather complex partition function $P(n)$.
In particular, $p(k,k)=P(k)$.
A: introduce k-1 dummy variables that would help u remove all the <= and covert them to < between all the 0<=x1----- then u can simply use the formula n+r-1Cr
