I am doing a couple of exercise questions, How do I show that if we let $n \geq 1$ be an integer, and if we consider $n$ people $P_1$,$P_2$,...,$P_n$. If we let $A_n$ be the number of ways these $n$ people can be divided into groups, such that each group consist of either one or two people.
How do we determine $A_{1}$, $A_{2}$, and , $A_{3}$ and how do we prove that for each integer $n\geq 3$,
$A_{n}$ = $A_{n-1}$ + $({n-1})$ $\cdot$ $A_{n-2}$
For $n=1$, there is only one grouping : ${P_1}$. That is, $A_1=1$. For $n=2$, there are two groupings: $\brace{\brace P_1, P_2}$ and $\brace {{\brace{P_1}}, {\brace {P_2}}}$. That is, $A_2 = 2$. Now let $n \geq 3$. Consider any grouping of $n$ persons such that each group contains either $1$ or $2$ persons. Take the group containing the last person, $P_n$. Either that group contains only $P_n$ or it contains one more person. In the first case, the remaining $n-1$ persons can be grouped in $A_{n-1}$ ways and in the second case, $P_n$'s group-mate could be any one of the remaining $n-1$; those two form a group and the remaining $n-2$ can be grouped in $A_{n-2}$ ways. That is, number of groupings in the second case $= (n-1) A_{n-2}$. Total number of groupings for $n$ persons, $$A_n=A_{n-1} + (n-1)A_{n-2}$$.