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let $n \gt 1$ be an integer, and consider n people; $P_1,P_2,...,P_n$ let $A_n$ be the number of ways these $n$ people can be divided into groups, such that each group have either one or two people

determine $A_1,A_2,A_3$ So I have

$A_1=1$ way

$A_2=2$ way

$A_3=4$ way

and from this I derive the formula $A_n = 2^{n-1}$ but this is not true for $n\ge 4$

so I have to prove that for each integer $n\ge 3$

$$A_n = A_{n-1} + (n-1) * A_{n-2}$$

So I proved this by Induction:

Where I assumed that $n \ge 2$ so $A_{n+1} = 2^n$

So I did the induction step.. and I was wrong

I was hoping anyone on here could help me by proving this with induction?

Thanks

Need Help
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    Your formula $A_n=2^{n-1}$ is only true for $n=1$, not $2$ or $3$ as well. – Ross Millikan Oct 13 '14 at 16:49
  • Your guess that $A_n=2^{n-1}$ only works for the $n=1$ case, since $2^{2-1}=2\neq4$ and $2^{3-1}=4\neq 12$. So you're trying to prove a formula that isn't valid. I'd suggest using your recurrence relation to derive a longer list of numbers so you have a better chance of seeing a pattern. – Semiclassical Oct 13 '14 at 16:50
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    To get a clearer understanding of what you intend, and how you are counting, it would be helpful to see examples of how you formed the groups. – Bill Province Oct 13 '14 at 16:50
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    Shouldn't $A_2=2$? Either two groups of $1$ or one group of $2$? And I think $A_3=4$ (three groups of $1$ or three different ways to have a group of $2$ and a group of $1$). – paw88789 Oct 13 '14 at 16:50
  • Do you want to prove this induction formula or you want to find a closed form $A_n$? – John Oct 13 '14 at 17:00
  • John, I want to prove this formula by induction and show it is true – Need Help Oct 13 '14 at 17:04
  • @NeedHelp You can argue this formula is true $\forall n\geq 3$ directly. I have writen down the details in my answer. – John Oct 13 '14 at 17:09
  • @JohnZHANG I know it can be proven by direct proof. But I also need to find a formula for An as well and the proof below does not find any formulas – Need Help Oct 13 '14 at 17:51

2 Answers2

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$A_n = A_{n-1} + (n-1)A_{n-2}$ can be shown by direct argument.

Case 1 from $A_{n-1}$ to $A_n$, we add 1 person, let this person forms a group alone, then we have $A_n$ arrangement. In total $A_{n-1}$.

Case2 from $A_{n-1}$ to $A_n$, we add 1 person, but we let this person form a group with another person from previous $n-1$ persons, so we have $n-1$ choices, the other $n-2$ persons have $A_{n-2}$ arrangement. In total $(n-1)A_{n-2}$.

John
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Such arrangements of $\{1, 2, \dotsc, n\}$ correspond in an obvious way to elements of $S_n$ that have order $2$; that is, involutions in $S_n$. Now this answer solves the problem; see also OEIS A000085.

There is a link from the article above to a Wikipedia page on the Telephone Numbers that provides some further exposition as well and discusses the recurrence you give. There does not appear to be a closed form.

Community
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rogerl
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