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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

- Proving $\sum_{k=1}^nk^3 = \left(\sum_{k=1}^n k\right)^2$ using complete induction
- A sum of fractional parts.
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- explanation for a combinatorial identity involving the binomial coefficient
- Proving $\phi(m)|\phi(n)$ whenever $m|n$

$A_1 = 1$ way

$A_2 = 4$ way

$A_3 = 12$ way

But I am not sure if this is right… Can anyone confirm?

edit:

My way of getting this is

$A_1 = 1 $ because {P1} only has 1 way to sort

$A_2 = 4$ because {P1,P2} = {P1,P2},{P2,P1},{P1P2},{P2P1}

etc.

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The only case that hasn’t been handled in the comments is $n=3$. If each group must have either one or two people, and you have three people altogether, there are two basic types of groupings: you can put each person into his own group, or you can have one group of two people and one singleton. The first possibility gives you the groups $\{P_1\},\{P_2\}$, and $\{P_3\}$. There are several ways to get a group of two and a singleton; can you see what they are?

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