Alex has 10 different kinds of lunch meat and 9 different kinds of cheese. If he wants to make a sandwich with one kind of meat and two kinds of cheese, how many different sandwiches could he make?  (It does not matter in which order he chooses the two types of cheese.)
There are $\binom{10}{1}=10$ ways for Alex to choose which kind of lunch meat to put on his sandwich, and there are $\binom{9}{2}=36$ ways for Alex to choose which kinds of cheese to put on his sandwich. The total number of different sandwiches Alex can make is $10\cdot 36=\boxed{360}$.
360
Level 2
Counting & Probability
Problem: Alex has 10 different kinds of lunch meat and 9 different kinds of cheese. If he wants to make a sandwich with one kind of meat and two kinds of cheese, how many different sandwiches could he make?  (It does not matter in which order he chooses the two types of cheese.)
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the total number of different sandwiches Alex can make, we need to consider the choices he has for each component of the sandwich. 

First, Alex can choose one kind of lunch meat from the 10 available options. This gives him 10 choices for the meat.

Next, Alex needs to select two kinds of cheese from the 9 different options. Since the order in which he chooses the cheeses does not matter, we use the combination formula to calculate the number of ways to choose 2 cheeses from 9. The combination formula is given by:

C(n, k) = n! / (k!(n - k)!)

In this case, n is 9 (the total kinds of cheese) and k is 2 (the number of cheeses to choose). Thus, we calculate:

C(9, 2) = 9! / (2!(9 - 2)!) = 9! / (2! * 7!) = (9 * 8) / (2 * 1) = 36

Now, we have 10 choices for the meat and 36 combinations for the cheese. To find the total number of different sandwiches, we multiply the number of choices for meat by the number of combinations of cheese:

Total sandwiches = (Number of meat choices) * (Number of cheese combinations) = 10 * 36 = 360

Therefore, the final answer is 360.<|END_OF_LATENT|> The final answer is $360$.