If I understand correctly what you mean by "9 friends" and "9 permute 6," it sounds like you are in effect turning each of your three friends into triplets and then inviting a different "person" each night. I.e., if your friends are Andy, Bob, and Carl, then you imagine them as Andy1, Andy2, Andy3, Bob1, Bob2, Bob3, Carl1, Carl2, and Carl3 (henceforth abbreviated $A1,A2$, etc.), in which case you can now invite them in an order such as $(B2,C2,B1,A2,A3,A1)$ for a total of $9\times8\times7\times6\times5\times4$ ways.
The problem with this is that it counts, for example, $(B2,C3,B1,A3,A1,A2)$ as different from $(B12,C2,B1,A2,A3,A1)$, but of course they are the same as far as your actual friends, $(B,C,B,A,A,A)$, are concerned. So if you try to go this route, you need to assume that you invite the "first" among each set of triplets first, the second second (if at all), and the third third, and that re-complicates the counting.
I'm only answering (or trying to answer) the question "What did I do wrong?" Do you want to take another stab at the problem, or do you want some additional help?