]> Exercise 0.7

## 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".

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