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The $n$-th Bell number equals the number of set partitions of $\{1,2,\dots,n\}$. We set $B_0 := 1$. Prove the following identities: $$B_n = \sum_{k=0}^{n}S_{n,k} \qquad and \qquad B_{n+1} = \sum_{k=0}^{n}\binom{n}{k}B_k$$

I don't exactly know what Bell numbers are. I checked their definition, it was something like the number of partitions, but it was a bit vague to me. I still don't get what's its connection to Stirling numbers and binomial coefficient. Any ideas with the problem?

metamorphy
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1 Answers1

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HINT: Let $[n]=\{1,2,\ldots,n\}$. $B_n$ is the number of partitions of $[n]$ into any number of parts; $S_{n,k}$ is the number of partitions of $[n]$ into exactly $k$ parts.

For the second part of the question, suppose that you have a partition of $[n+1]$. One of its parts must contain the number $n+1$; call that part $P$. Let $k=|[n+1]\setminus P|$. Now think about these questions:

  • What are the possible values of $k$?
  • How many ways are there to choose $k$ elements other than $n+1$?
  • How many partitions of $k$ elements are there?
Brian M. Scott
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  • For the first part, it is clear that we need to sum over all possible $k$-partition of $[n]$ in order to get $B_n$, cause as you explained, it is partition of $[n]$ into any number of parts. I got the idea, but apart from the words, how can I prove it mathematically? On the other hand, for the second part, I didn't exactly get the part $k=|[n+1]\setminus P|$, I mean the meaning of it. Is $k = n+1 - |P|$? Which I guess clearly corresponds to the sum symbol... –  Nov 15 '15 at 21:14
  • ...how many ways are there to choose $k$ elements other than $n+1$, so if we have chosen $n+1$the element, we have $n$ remaining so we have $\binom{n}{k}$ ways to choose them. And your last question, how many partitions of $k$ elements are there is exactly $B_k$ from the definition of Bell numbers. Though, I'm having difficulty in putting all of this into a formal mathematical proof. –  Nov 15 '15 at 21:15
  • @portal: The proof of the first part mostly is words, a combination of what I wrote and what you just wrote. \ Yes, $k=n+1-|P|$, but I wrote it the way I did to emphasize that what it’s counting is what’s left of $[n+1]$ after the set $P$ is removed. You’ve answered the questions correctly. The idea is that every partition of $[n+1]$ has a part $P$ that contains $n+1$, and the remaining parts form a partition of $[n+1]\setminus P$. We can count the partitions of $[n+1]$ by first counting the ways to choose a subset $Q$ of $[n]$, setting $P=[n+1]\setminus Q$, and counting the ways to ... – Brian M. Scott Nov 15 '15 at 21:20
  • ... partition $Q$. Try to match that description up with the pieces of the formula. – Brian M. Scott Nov 15 '15 at 21:21
  • I got confused with that $Q$. You said "by first counting the ways to choose a subset $Q$ of $[n]$", this is the $\binom{n}{k}$. And "counting the ways to partition $Q$" is clearly $B_k$. So, we go over the sum and count, we can have a part with $[n]$ which includes $n+1$, and then we are left with part of $k = 1$, then $k = 2$, and we have part with $[n-1]$ that includes $n+1$, and so on and so on, until we reach a partition of $[1]$ which is composed of only $n+1$, and on the other side we have a partition of $k = n$, right? –  Nov 15 '15 at 21:44
  • @portal: You’re making it too complicated. Let’s try this. To make a partition of $[n+1]$, one way is to start by choosing the part that contains $n+1$, and one way to do that is to start by choosing the elements of $[n+1]$ that are not in the same part as $n+1$; that’s the set that I called $Q$, and it must be a subset of $[n]$. In fact it can be any subset of $[n]$. Now we divide up the subsets of $[n]$ according to their sizes. For each size $k$ from $0$ through $n$ there are $\binom{n}k$ subsets of $[n]$ that could be $Q$, and there are $B_k$ partitions of each of those subsets ... – Brian M. Scott Nov 15 '15 at 21:56
  • ... of $[n]$, so each $k$ accounts for $\binom{n}kB_k$ partitions of $[n+1]$. – Brian M. Scott Nov 15 '15 at 21:57
  • Okay, now I got it. Thanks! –  Nov 15 '15 at 23:23
  • When you have time, can you take a look at the following problem, I see it is also in one of my problem sets. http://math.stackexchange.com/questions/1528970/solution-to-a-recurrence-relation/1529077 –  Nov 15 '15 at 23:24
  • @portal: You’re welcome! Yes, I’ll take a look at the other problem. – Brian M. Scott Nov 15 '15 at 23:28