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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".

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