Let's use smaller numbers. Let's say there are only 3 people, and we want to count the number of ways to put them in a group of 2 and a group of 1. According to your method, this is

Suppose the people are A, B, and C. Let's list off each possible grouping:

{A,B} and {C}

{A,C} and {B}

{B,C} and {A}

We could also permute the two people in each group, but we get the same group. The order doesn't matter. So these are the only three. See if you can figure out why your counting method overcounted.

Back to the original problem: let's look at permutations of the nine people. The first six people (in the permutation) go to one group and the last three go to another. For a given permutation, let's say [A,B,C,D,E,F,G,H,I], we obtain the groups {A,B,C,D,E,F} and {G,H,I}. However, any permutation where the letters A through F appear in the first six positions and G through I appear in the last three gives this same grouping. So, among those permutations, we get this same grouping how many times? It is the number of ways to permute 6 elements times the number of ways to permute 3 elements. Since this is true for EVERY pair of groups, we are counting the groups times each. Hence, the number of groups is