How do I solve a probability problem involving permutations and having two steps?

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.

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.