1. **Stating the problem:** Construct a real-world situation involving counting techniques and probability, identify the techniques to use, and solve the problem. 2. **Situation:** Imagine a small bakery that offers 3 types of bread (white, wheat, rye) and 4 types of fillings (ham, cheese, tomato, lettuce). A customer wants to buy a sandwich with exactly one type of bread and two different fillings. 3. **What counting techniques and probability to use?** We use combinations to count the number of ways to choose fillings because the order of fillings does not matter. We use the multiplication principle to combine choices of bread and fillings. 4. **Solution and Answer:** - Number of ways to choose 1 bread from 3 types: $3$ - Number of ways to choose 2 fillings from 4 types: $\binom{4}{2} = \frac{4!}{2!\times(4-2)!} = \frac{24}{2\times2} = 6$ - Total number of different sandwiches: $3 \times 6 = 18$ Therefore, there are **18** different possible sandwiches. If a customer randomly picks one sandwich from all possible sandwiches, the probability of choosing any particular sandwich is $\frac{1}{18}$. This problem uses the combination formula $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ and the multiplication principle for counting.