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Consider splitting an integer interval $[1..N]$ into non-overlapping subintervals $[i_k..j_k],\,i_k \le j_k$, i.e. $$ [1..N] = \bigcup_{k=1}^{K} \, [i_k..j_k] \quad i_1 =1, i_{k+1} = j_{k}+1, j_K=N $$ How many ways can this be done?

Example: $[1,2,3]$: The set of possibilities is $\{[1]\cup[2]\cup[3],[1]\cup[2,3],[1,2]\cup[3],[1,2,3] \}$ for a total of $4$.

This question is related to Splitting the set $A=\{1,2,...,n\}$ into at most $m$ non-empty disjoint subsets, whose union is $A$, but differs in that here the subsets must consist of contiguous numbers.

BeMuSeD
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2 Answers2

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The wanted number is \begin{align*} \color{blue}{2^{N-1}} \end{align*} We can identify each of the intervals with the left-most value as in the case $N=3$: \begin{align*} \{[\color{blue}{1}]\cup[\color{blue}{2}]\cup[\color{blue}{3}],[\color{blue}{1}]\cup[\color{blue}{2},3],[\color{blue}{1},2]\cup[\color{blue}{3}],[\color{blue}{1},2,3] \}\qquad\longleftrightarrow\qquad (1,2,3),(1,2),(1,3),(1) \end{align*}

We can write this number as sum \begin{align*}\color{blue}{\sum_{j=1}^N\sum_{1=i_1<i_2<\cdots<i_j\leq N}1}&=\sum_{j=1}^{N}\binom{N-1}{j-1}\tag{1}\\ &=\sum_{j=0}^{N-1}\binom{N-1}{j}\tag{2}\\ &\,\,\color{blue}{=2^{N-1}} \end{align*} according to the claim.

Comment:

  • In (1) we note that $j_1=1$ since we always start with $1$. $\binom{N-1}{j-1}$ is the number of $j-1$-tuples $\left(i_2,i_3,\ldots,i_{j}\right)$ between $2$ and $N$.

  • In (2) we shift the index by one to start with $j=0$ and apply the binomial theorem.

Markus Scheuer
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The number of ways is $2^{n-1}$. Imagine $n$ objects in a row, numbered $1$ to $n$ from left to right. There are $n-1$ gaps between adjacent objects. In each gap, you can either place a barrier, or not. You can make these $n-1$ binary choices is $2^{n-1}$ ways. The placement of the barriers uniquely determines a partition of $\{1,\dots,n\}$ into intervals.

Mike Earnest
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