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Difficulty: MediumCombinations and Group Selections

A restaurant culinary team is designing a tasting menu consisting of 44 distinct dishes selected from a pool of 55 appetizers and 66 main courses. The tasting menu must contain at least 11 appetizer and at least 22 main courses. Furthermore, a specific appetizer, Dish A, can only be included on the menu if a specific main course, Dish B, is also included. How many different valid tasting menus can be formed?

  1. A
    180
  2. 200Answer
  3. C
    240
  4. D
    250
  5. E
    310

Answer

200 different valid tasting menus can be formed.
The total number of valid tasting menus is 200200. This is calculated by breaking down the total requirement into two mutually exclusive dish distributions: 1 appetizer with 3 main courses (giving 9090 valid menus) and 2 appetizers with 2 main courses (giving 110110 valid menus), accounting for the restriction that selecting Dish A requires Dish B.

Step-by-Step Solution

1
Determine the valid distribution of dishes between appetizers and main courses.
Since 44 dishes must be selected with at least 11 appetizer and at least 22 main courses, the only possible distributions are: Case 1 (11 appetizer, 33 main courses) and Case 2 (22 appetizers, 22 main courses).
Selecting 33 appetizers and 11 main course violates the constraint of having at least 22 main courses.
2
Calculate the valid combinations for Case 1 (1 appetizer and 3 main courses).
Subcase 1a (Dish A not selected): Choose 11 appetizer from the remaining 44 appetizers and 33 main courses from all 66 main courses: (41)×(63)=4×20=80\binom{4}{1} \times \binom{6}{3} = 4 \times 20 = 80.
Subcase 1b (Dish A selected): Dish B must also be selected. Choose no additional appetizers from the remaining 44, and choose 22 more main courses from the remaining 55 main courses: 1×(52)=101 \times \binom{5}{2} = 10.
Total for Case 1: 80+10=9080 + 10 = 90.
Separating into cases based on whether Dish A is included properly enforces the conditional constraint.
3
Calculate the valid combinations for Case 2 (2 appetizers and 2 main courses).
Subcase 2a (Dish A not selected): Choose 22 appetizers from the remaining 44 appetizers and 22 main courses from all 66 main courses: (42)×(62)=6×15=90\binom{4}{2} \times \binom{6}{2} = 6 \times 15 = 90.
Subcase 2b (Dish A selected): Dish B must also be selected. Choose 11 more appetizer from the remaining 44 appetizers and 11 more main course from the remaining 55 main courses: (41)×(51)=4×5=20\binom{4}{1} \times \binom{5}{1} = 4 \times 5 = 20.
Total for Case 2: 90+20=11090 + 20 = 110.
Evaluates all valid menus having 22 appetizers and 22 main courses.
4
Sum the valid combinations from both cases.
Total valid menus = 90+110=20090 + 110 = 200.
The two main distribution cases are mutually exclusive.

Key Concept

Combinations with multiple category constraints and conditional dependencies
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Topics

Combinations and Group Selections

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