How do I solve a probability problem involving permutations and having two steps? 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
 A: 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.
A: If the man starts the first night eating in a Chinese restaurant, he has 9 choices. In the second night, for an Italian restaurant, he has 8 choices.
In the 3rd night, he has 8 Chinese restaurants left, to eat at. And so on.
This gives: $9\cdot 8\cdot 8\cdot 7\cdot 7\cdot 6\cdot 6$
If he starts eating in an Italian restaurant, he has 
$8\cdot 9\cdot 7\cdot 8\cdot 6\cdot 7\cdot 5$
We have to add these to know how many possible ways there are to spend his 7 days eating from different restaurants each night.
