TITLE: How do I solve a probability problem involving permutations and having two steps?
QUESTION [6 upvotes]: The problem: 

A man spends 7 nights in a city. 
  He has a list of the 8 best Italian restaurants and the 9 best Chinese restaurants. How many ways can he eat 7 meals at these restaurants, assuming a different restaurant each night and he wishes to alternate between Italian and Chinese food.

I first tried using permutations using $n=17$ and $r=7$. The result: $98.017.920$. Then knowing I had to do something with alternating the restaurants and assuming that he starts with an Italian restaurant, there would be 4 Italian and 3 Chinese restaurants. So I used permutations for Ital. $n=8,r=4$ and Chin. $n=9,r=3$ for $1680$ and $504$ respectively. I divided the product but I know that is not the answer. I don't have a good grasp of when to use permutations or combinations.
Any help in clearing this up would be greatly appreciated.
Thank you

REPLY [2 votes]: I suggest another solution:

Assuming he starts with an Italian restaurant, thus having four Italian and three Chinese meals: $\binom84\cdot\binom93=5880$ possibilities.
Assuming he starts with a Chinese restaurant, it's $\binom94\cdot\binom83=7056$ possibilities.

This yields a total of 12936 possibilities.
My reasoning is that the days (or nights) do not matter: As suggested by @JimM in his comment, Italian_1 on the first night and Italian_2 on the second is the same result as Italian_2 on the first and Italian_1 on the second -- in both ways, he eats at those two restaurants.