How to find the restricted partition of n into k *distincts* parts between a finite set [1;r]? It seems to be an opened question.
Indeed, it is easy to find:


*

*the number of partitions of n into k distinct parts 

*the number of partitions of n into k parts

*the number of partitions of n into k parts, each of them less or equal than r and thus the k parts between a finite set [0;r].


However, I have not found any formula for partitions with three restrictions: k parts, k must be distinct, and k must belong to S (S=[1;r]). 
To be concrete, what is the number of partition of 60 into 5 distinct parts all of them between [1;50] ?
 A: Actually in your particular example you can remove the "all of them between [1;50]" requirement since any partition of $60$ into $5$ distinct positive parts cannot have a part greater than $50$ (the other four must be at least $1+2+3+4$).  In your particular example, the answer is $2611$.
When Java applets worked on the internet, I would have pointed you at my page http://www.se16.info/js/partitions.htm which does the calculation by recursion. 
The number of partitions of $n$ into exactly $k$ distinct positive parts each no greater than $r$ is same as the number of partitions of $n-\frac{k(k+1)}{2}$ into up to $k$ (not necessarily distinct) positive parts each no greater than $r-k$: you can subtract $1,2,\ldots,k$ from the smallest, second smallest, ..., largest terms respectively.  So in your example the number of partitions of $45$ into up to $5$ positive parts each no greater than $45$ (you see again the point about an unnecessary constraint). 
Let's use $f(x,y,z)$ for the number of partitions of $x$ into up to $y$ (not necessarily distinct) positive parts each no greater than $z$.  You start with a lot of zeros but $f(0,0,0)=1$ and then use $$f(x,y,z)=f(x,y,z-1)+f(x-z,y-1,z).$$ 
