Question

The number of different ways in which the first twelve natural numbers can be divided into three groups, each having four elements, such that the numbers in each group are in arithmetic progression is

Answer 1

No group can have a common difference $\geq 4$ as there is no number $\geq 13$
If the group containing $1$ has the common difference $=1$ i.e. if one group is $\left(1,2 , 3,4\right)$ , there will be two ways to group others - $\left(5,6 , 7,8\right)$ , $\left(9,10,11,12\right)$ and $\left(5,7 , 9,11\right),\left(6,8 , 10,12\right)$ .
If the group containing $1$ has the common difference $=2$ i.e. one group is $\left(1,3 , 5,7\right)$ , there will be one way to group others $\left(2,4 , 6,8\right),\left(9,10,11,12\right)$ .
If the group containing $1$ has common difference $=3$ i.e. if one group is $\left(1,4 , 7,10\right)$ there will be one way to group others $\left(2 , 5 , 8 , 11\right),\left(3 , 6 , 9 , 12\right)$
$\Rightarrow $ The number of ways $=2+1+1=4$