Permutations and combinations: Problem type 3 A catering service offers 6 appetizers, 10 main courses, and 3 desserts. A customer is to select 5 appetizers, 6 main courses, and 2 desserts for a banquet. In how many ways can this be done?

Step 1 ：Translate the problem into a combination problem. The formula for combinations is \(C(n, k) = \frac{n!}{k!(n-k)!}\), where n is the total number of items, k is the number of items to select, and ! denotes factorial.

Step 2 ：Calculate the number of combinations for each course separately. For appetizers, \(C(6, 5) = \frac{6!}{5!(6-5)!} = 6\). For main courses, \(C(10, 6) = \frac{10!}{6!(10-6)!} = 210\). For desserts, \(C(3, 2) = \frac{3!}{2!(3-2)!} = 3\).

Step 3 ：Multiply the number of combinations for each course together to get the total number of ways to select the banquet menu. \(6 * 210 * 3 = 3780\).

Step 4 ：Final Answer: The total number of ways to select the banquet menu is \(\boxed{3780}\).