jmc

First, we need to find the sum of each group when split. This is the total sum of all the elements divided by the # of groups. $1+2\cdots +9=\frac{9(10)}{2}=45$. Then, dividing by $3$, we have $\frac{45}{3}=15$, so each group of $3$ must have a sum of 15. To make the counting easier, we will just see the possible groups 9 can be with. The possible groups 9 can be with 2 distinct numbers are $(9,2,4)$ and $(9,1,5)$. Going down each of these avenues, we will repeat the same process for $8$ using the remaining elements in the list. Where there is only 1 set of elements getting the sum of $7$, $8$ needs in both cases. After $8$ is decided, the remaining 3 elements are forced in a group, yielding us an answer of $\boxed{2}$ as our sets are $(9,1,5)(8,3,4)(7,2,6)$ and $(9,2,4)(8,1,6)(7,3,5)$