Okay, let's try to figure out how many different meal combinations Yann and Camille can order if they don't want to have the same dish. The restaurant has 10 items on the menu, and each of them orders one dish. Also, it matters who orders what, so the order matters here—like Yann ordering chicken and Camille ordering fish is different from Yann ordering fish and Camille ordering chicken. 

Hmm, so first, let me think. If there were no restrictions, meaning they could order the same dish, how many combinations would there be? Well, Yann has 10 choices, and Camille also has 10 choices, so that would be 10 multiplied by 10, which is 100. But since they refuse to order the same dish, we have to subtract the cases where they ordered the same dish. 

Wait, but maybe there's another way to think about it. Since the order matters, maybe we can approach this as permutations instead of combinations. Let's break it down. 

For Yann, he can choose any of the 10 dishes. Once he's chosen his dish, Camille can't choose the same one. So, Camille has 9 remaining choices. Therefore, the total number of combinations would be 10 (for Yann) multiplied by 9 (for Camille), which is 90. 

But let me verify that. Let's say there are 10 dishes: A, B, C, D, E, F, G, H, I, J. If Yann picks A, Camille can pick B through J—9 options. Similarly, if Yann picks B, Camille picks A, C through J—again 9. So, yes, each of Yann's 10 choices gives Camille 9, so total 90. 

Alternatively, if we considered combinations without order, it would be different. But the question specifies that the order matters. So, combinations where Yann and Camille switch dishes are considered different. Therefore, each permutation of two different dishes counts as two different combinations. 

Another way: the number of permutations of 10 dishes taken 2 at a time. The formula for permutations is P(n, k) = n! / (n - k)!. So here, n=10, k=2. So P(10, 2) = 10! / (10-2)! = 10! / 8! = 10 × 9 × 8! / 8! = 10 × 9 = 90. 

Yes, that matches the previous result. So both methods give 90. 

Wait, but let's make sure there's no overlap or something I'm missing. For example, if there are 10 dishes, and each person picks one, with no repeats. So the first person has 10 options, the second has 9. So 10*9=90. That seems straightforward. 

Alternatively, if they could choose the same dish, it's 10*10=100, but since they can't, subtract the cases where they chose the same dish. How many cases where they chose the same dish? For each dish, there's one case where both chose that dish. There are 10 dishes, so 10 cases where they ordered the same dish. Therefore, total combinations where they didn't order the same dish would be 100 - 10 = 90. 

Yes, that also gives 90. So both approaches confirm the answer is 90. 

Therefore, I think the answer is 90 different combinations.
</think>