Exercise 0.7

]> Exercise 0.7

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Exercise 0.7

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 $n - 1$ element set obtained by ignoring $n$ .

and those that contain $n$ : these all have $k - 1$ elements of the remaining $n - 1$ .

This gives the recursion

C (n, k) = C (n - 1, k) + C (n - 1, k - 1)

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 $k - 1$ 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 $n - 1$ element set into $k$ blocks) different ways that this second possibility can happen.

This gives the recursion used in the spreadsheet here, which is

S (n, k) = S (n - 1, k - 1) + k * S (n - 1, k)

which can be read as " $n$ stands alone, or with one of the $k$ blocks that exist without it".