Partition a set of m objects into n subsets of positive size and no two of them should have equal size. In how many ways can I partition a set of size m into n subsets such that each of them must have atleast one element and no two of them should have same no. of elements. ? 
 A: I had a Java applet which calculated these sorts of things at http://www.se16.info/js/partitions.htm though with worries about Java security, most modern browsers will not allow you to use it.  The source code is linked there too.
You can calculate the answer with a simple recurrence, since you can construct a partition of $m$ with $n$ distinct positive parts:


*

*either by taking a partition of $m-n$ with $n$ distinct positive parts and adding $1$ to each part

*or by taking a partition of $m-n$ with $n-1$ distinct positive parts, adding $1$ to each part, and then adding a part of size $1$


so the solution satisfies $$f(m,n) = f(m-n,n-1) + f(m-n,n)$$ starting from $f(0, 0)=1$, $f(m,0)=0$ if $m \gt 0$, and $f(m,n)=0$ if $m \lt 0$. 
Another approach can be derived from considering the number of ways of partitioning the positive integer $m$ into $n$ distinct positive integers is the same as the number of ways of partitioning $m- \frac{n(n+1)}{2}$  into up to $n$ positive integers (not necessarily distinct), which is also the number of ways of partitioning $m- \frac{n(n+1)}{2}$  into positive integers where the largest is no greater than $n$.  As an example, $7$ can be partitioned into $2$ distinct positive parts in three ways: $$7 = 4+3 = 5+2=6+1$$
corresponding to $7-\frac{2\times (2+1)}{2}=4$ being partitioned into up to $2$ positive integers in three ways: $$4 = 2+2 = 3+1 = 4 (+0)$$
and corresponding to $4$ being partitioned into positive integers each no more than $2$ in three ways: $$4=2+2 = 2+1+1 = 1+1+1+1.$$
OEIS A060016 and A008289 have more details.
