Let the universe S be the natural numbers between 1 and N. And let's say we have K consecutive subsets $A_1 \subseteq A_2 \subseteq \ldots \subseteq A_K \subseteq S $, where all the subsets $$A_i$$ must be a subset of $$A_i+1$$ and subset $$A_K$$ must be a subset of the universe S.

Note: the subsets do not need to be proper subsets.

If we are given N and K, how many different subset chains are there?

Example 1: N = 2, K = 1

There are 4 ways to make 1 valid subset chain.

{} (selecting the empty set)




Example 2: N = 1, K = 2

There are 3 ways to make 2 valid subset chains.




Example 3: N = 2, K = 2




{},{1, 2}



{2}, {1,2}

{2}, {2}



A solution $A_1 \subseteq A_2 \subseteq \ldots \subseteq A_K \subseteq S $ corresponds to a partition of $S$ into $K+1$ disjoint subsets $A_1, A_2 \backslash A_1, \ldots, A_K \backslash A_{K-1}, S \backslash A_K$. The number of these is easily seen to be $(K+1)^N$, as each member of $S$ can go into any one of the $K+1$ subsets.

  • $\begingroup$ This question is not well worded, I think the OP also admits subsets that are equal as opposed to proper subsets. Also it is not clear that OP uses all $N$ values. $\endgroup$ – Marko Riedel Sep 26 '16 at 21:13
  • $\begingroup$ @MarkoRiedel I'm not assuming proper subsets, and I'm not assuming using all $N$ values in the $A_i$ (that's why there's the $S \backslash A_K$). $\endgroup$ – Robert Israel Sep 26 '16 at 21:28
  • $\begingroup$ I apologize if the wording is not great, I don't know the math notation very well. @RobertIsrael is correct in assuming they are not proper subsets. $\endgroup$ – mpgaillard Sep 26 '16 at 22:10
  • $\begingroup$ @RobertIsrael Thank you for explaining the answer. (+1). $\endgroup$ – Marko Riedel Sep 26 '16 at 22:13
  • $\begingroup$ Awesome. Thanks! $\endgroup$ – mpgaillard Sep 26 '16 at 22:42

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