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
Best Answer
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.