How many ways are there to invite one of three different friends over for dinner on six successive nights such that no friend is invited more than three times?

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Okay so here's the full problem followed what I'm thinking:

Problem: How many successive ways are there to invite one of four different friends over for dinner on five successive night s.t no friend is invite more than three times?

Now here's the though process I'm going down:

We proceed by cases, adding up the cases at the end

Case 1: 1 friend for 3 nights, 1 friend for 2 nights

Let's denote the friends as $f_i$ where $1\leq i\leq 4$. If $f_1$ eats dinner at the house for 3 nights, that leaves either $f_2$ for two nights, or $f_3$ or $f_4$. Thus, there are 3 arrangements if $f_1$ eats dinner 3 nights. It follows similarly if we consider $f_2$ for the spot of 3 nights, or $f_3$ or $f_4$. Thus, there are $$12{5 \choose 3}{2 \choose 2}$$ arrangements possible.

Case 2: 1 friend for 3 nights, 1 friend for 1 night, 1 friend for 1 night

If $f_1$ eats dinner at the house 3 nights, we have $f_2$ one night, $f_3$ one night. Or we have $f_2$ one night, $f_4$ one night. Or we have $f_3$ one night, $f_4$ one night. Thus, with $f_1$ on 3 nights, we have 3 ways to choose. The subcases of $f_2-f_4$ follow similiarly, so we have $$12{5 \choose 3}{2 \choose 1}{1 \choose 1}$$

Am I on the right track or does there need to be some revision?