A restaurant offers a dinner special for two people for $20 that lets you choose from 9 entrees, 10 side dishes, and 5 desserts. You can choose one entree, two side dishes, and two desserts. However, you're not permitted to order two of the same sides, or two of the same desserts. How many meals are possible?
A restaurant offers a dinner special for two people for $20 that lets you choose from 9 entrees, 10 side dishes, and 5 desserts. You can choose one entree, two side dishes, and two desserts. However, you're not permitted to order two of the same sides, or two of the same desserts. How many meals are possible?
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Restaurant Dinner Special Combinatorics Problem**
A restaurant offers a dinner special for two people for $20 that lets you choose from 9 entrees, 10 side dishes, and 5 desserts. You can choose one entree, two side dishes, and two desserts. However, you're not permitted to order two of the same sides or two of the same desserts. How many meals are possible?
This problem can be framed in the context of combinatorics and probability for educational purposes. Students can learn to apply counting principles and combinations to solve real-world problems.
**Explanation:**
1. **Entrees:** There are 9 options for entrees. From the principle of counting, the number of ways to choose 1 entree from 9 is 9.
2. **Side Dishes:** There are 10 options for side dishes, and you need to choose 2 different ones. This is a combination problem where order does not matter.
- The number of combinations of 10 items taken 2 at a time is calculated using the combination formula \( \binom{n}{k} = \frac{n!}{k! (n-k)!} \):
\[
\binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45
\]
3. **Desserts:** Similarly, there are 5 options for desserts, and you need to choose 2 different ones.
- The number of combinations of 5 items taken 2 at a time:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
4. **Total Number of Possible Meals:** To find the total number of different possible meals, multiply the number of choices for each component:
\[
9 \times 45 \times 10 = 4050
\]
Thus, the total number of different meal combinations possible is **4050**. This showcases the application of combinatorial methods in solving real-world selection problems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F95e5613b-1a25-4b20-8027-b25d1244d9dc%2F3c945023-daa7-481d-9ac1-b5037b27a400%2F1mavmee_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Restaurant Dinner Special Combinatorics Problem**
A restaurant offers a dinner special for two people for $20 that lets you choose from 9 entrees, 10 side dishes, and 5 desserts. You can choose one entree, two side dishes, and two desserts. However, you're not permitted to order two of the same sides or two of the same desserts. How many meals are possible?
This problem can be framed in the context of combinatorics and probability for educational purposes. Students can learn to apply counting principles and combinations to solve real-world problems.
**Explanation:**
1. **Entrees:** There are 9 options for entrees. From the principle of counting, the number of ways to choose 1 entree from 9 is 9.
2. **Side Dishes:** There are 10 options for side dishes, and you need to choose 2 different ones. This is a combination problem where order does not matter.
- The number of combinations of 10 items taken 2 at a time is calculated using the combination formula \( \binom{n}{k} = \frac{n!}{k! (n-k)!} \):
\[
\binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45
\]
3. **Desserts:** Similarly, there are 5 options for desserts, and you need to choose 2 different ones.
- The number of combinations of 5 items taken 2 at a time:
\[
\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10
\]
4. **Total Number of Possible Meals:** To find the total number of different possible meals, multiply the number of choices for each component:
\[
9 \times 45 \times 10 = 4050
\]
Thus, the total number of different meal combinations possible is **4050**. This showcases the application of combinatorial methods in solving real-world selection problems.
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