Home  18.013A  Chapter 0  Section 0.4 


Binomial coefficients count the number of subsets of an $n$ element set having $k$ elements in them. The Stirling number here counts the number of partitions of a set of $n$ elements into $k$ disjoint blocks. Prove these two statements.
Solution:
The subsets of an $n$ element set of size $k$ are of two kinds.
Those which do not contain the nth element: these are actually $k$ element subsets of an the $n1$ element set obtained by ignoring $n$ .
and those that contain $n$ : these all have $k1$ elements of the remaining $n1$ .
This gives the recursion
which is what is computed in the first spreadsheets.
A partition of an $n$ element set into $k$ blocks can also be of either of two kinds. In one the element $n$ is all by itself in a block; and we have a partition of the rest into $k1$ blocks.
Otherwise, it comes from a partition of the rest into $k$ blocks, and then it can go into any one of them; so there are $k$ * (the number of partitions of an $n1$ element set into $k$ blocks) different ways that this second possibility can happen.
This gives the recursion used in the spreadsheet here, which is
which can be read as " $n$ stands alone, or with one of the $k$ blocks that exist without it".
