1

Let $S$ be a set with $100$ elements. Divide $100$ by $20$. That leaves a partition of $S$ with $5$ subsets.

Suppose we need to choose a number of $k$-subsets from an $n$-set.

I'll try and reason by analogy with the partitioning of the set $S$ above. First, we calculate all the $k$-permutations. Let all these $k$-permutations be the elements of some set $A$. Then we partition $A$ into equivalence classes based on the relation that says all the permutations whose elements would make a single set are an equivalence class. Then all these equivalence classes will make up the subsets we are after.

Does that make sense?

effofexx
  • 35
  • You start with an example, but do not say whether $k$ corresponds to $5$ or $20$. Are you trying to figure out how many ways there are to partition $100$ into $20$ disjoint $5$ element subsets, or $5$ disjoint $20$ element subsets, or something else? – Ross Millikan Aug 04 '17 at 14:53

0 Answers0