Chapter 7 Set Partitions
The Stirling number of second kind \(S(n,k)\) is the number of partitions of the set \([n]\) into exactly \(k\) blocks. (A block is a nonempty set.) Here are the first rows of the Stirling triangle:
Exercises Practice Problems
1. Set Partitions into Two Blocks.
Let \(n\) be a positive integer. Find a general formula for \(S(n,2)\text{,}\) the number of ways to partition the set \([n]\) into two blocks.
2. The Next Row of the Stirling Triangle.
Use the Stirling Recurrence
to find the \(n=6\) row of the Stirling triangle.
3. Balls and Boxes.
What balls-in-boxes problem is equivalent to partitions of \([n]\) into \(k\) parts? There are four decisions to make about the balls and boxes.
4. Another Stirling Recurrence.
Stirling numbers of the second kind satisfy the following identity \[ S(n+1,k) = \sum_{i=0}^n { n \choose i} S(n-i, k-1). \]
Use the Stirling triangle we made in class to verify the identity is true for \(n=4\) and \(k=3\text{.}\) Note that \(S(n,0)=0\) and \(S(n,k)=0\) when \(k>n\text{.}\)
Give a combinatorial proof of the identity. That is, explain why the LHS and the RHS count the exact same thing.
5. Sets of Functions.
Consider the set \({\cal F}\) of all functions \(f : [n] \rightarrow [k]\text{.}\) For each subset of \({\cal F}\) below, determine (i) which balls-in-boxes problem is equivalent to the subset (ii) any restrictions on \(n\) and \(k\text{,}\) and (iii) the formula for the size of the set (if we know it).
The set of all functions \(f: [n] \rightarrow [k]\)
The set of injections \(f: [n] \rightarrow [k]\)
The set of surjections \(f: [n] \rightarrow [k]\)
The set of bijections \(f: [n] \rightarrow [k]\)
6. Writing \(x^n\) as a combination of \((x)_k\)'s.
In this problem, you will prove the following cool formula for \(n \geq 0\text{:}\)
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Recall that the set \(\R[x]\) of (finite) polynomials in \(x\) form a vector space over \(\R\text{.}\) Our typical basis for \(\R[x]\) is:
\begin{equation*} 1, x, x^2, \ldots, x^k, \ldots. \end{equation*}Explain why another basis for \(\R[x]\) is:
\begin{equation*} 1, (x)_1, (x)_2, (x)_3, \ldots, (x)_k, \ldots. \end{equation*}Just give an intuitive explanation (not a rigorous proof).
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Give a combinatorial proof that
\begin{equation*} m^n = \sum_{k=0}^n S(n,k) (m)_k. \end{equation*}In particular, both sides count the number of ways to assign \(n\) students to \(m\) classrooms where some classrooms can be empty. Explain.
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Fixing \(n\text{,}\) now consider the polynomial equation
\begin{equation*} x^n - \sum_{k=0}^n S(n,k) (x)_k= 0. \end{equation*}Explain why part (b) shows that this equation has an infinite number of solutions. Then conclude that the LHS must in fact be the zero polynomial, so
\begin{equation*} x^n = \sum_{k=0}^n S(n,k) (x)_k. \end{equation*}