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Difficulty: HardBasic Probability and Counting Methods

A high school cafeteria offers a lunch meal combination consisting of 1 main dish, 1 side dish, and 1 drink. There are 4 main dish options, 5 side dish options, and 3 drink options available. However, due to dietary restrictions, 1 specific main dish option cannot be paired with 2 specific side dish options. How many different valid lunch meal combinations of 1 main dish, 1 side dish, and 1 drink can a student choose?

Answer: 54 combinations

Answer

The total number of valid lunch meal combinations is 54.
The total number of unrestricted meal options is 4×5×3=604 \times 5 \times 3 = 60. The restricted main dish cannot be paired with 2 specific side dishes across any of the 3 drink options, creating 1×2×3=61 \times 2 \times 3 = 6 invalid combinations. Subtracting these from the total gives 60−6=5460 - 6 = 54 valid combinations.

Step-by-Step Solution

1
Calculate the total number of unrestricted meal choices
4×5×3=604 \times 5 \times 3 = 60 total combinations
By the Fundamental Counting Principle, multiplying the available choices for each dish component yields the total possible combinations without restrictions.
2
Determine the number of invalid combinations caused by dietary restrictions
1×2×3=61 \times 2 \times 3 = 6 invalid combinations
The 1 restricted main dish cannot be paired with 2 specific side dishes, and each of those 2 pairs can still be matched with any of the 3 drinks.
3
Subtract invalid combinations from the unrestricted total
60−6=5460 - 6 = 54 valid combinations
Subtracting the disallowed combinations leaves only the combinations that satisfy all dietary restrictions.

Key Concept

Fundamental Counting Principle with Restrictions
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Topics

Basic Probability and Counting Methods

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