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Difficulty: MediumPermutations and Linear Arrangements

A chef is arranging 6 distinct dishes—3 appetizers and 3 main courses—in a single straight row on a serving table for a culinary presentation. If all 3 appetizers must be placed adjacent to one another and all 3 main courses must also be placed adjacent to one another, in how many different linear arrangements can the dishes be displayed?

  1. A
    18
  2. B
    36
  3. 72Answer
  4. D
    144
  5. E
    720

Answer

72
To find the number of valid linear arrangements, treat the 3 appetizers as a single group and the 3 main courses as another single group. There are 2!=22! = 2 ways to arrange these two groups on the table (appetizers first or main courses first). Within the appetizer group, the 3 distinct dishes can be arranged in 3!=63! = 6 ways. Within the main course group, the 3 distinct dishes can also be arranged in 3!=63! = 6 ways. By the Fundamental Counting Principle, the total number of linear arrangements is 2!×3!×3!=2×6×6=722! \times 3! \times 3! = 2 \times 6 \times 6 = 72.

Step-by-Step Solution

1
Group the dishes into distinct blocks based on the given adjacency restriction.
Create Block A (containing all 3 appetizers) and Block M (containing all 3 main courses).
Because items of the same category must remain adjacent, each category functions as a single compound element.
2
Calculate the number of ways to arrange the 2 blocks relative to each other.
Number of block arrangements = 2!=22! = 2.
The two blocks can be placed in order (Block A then Block M) or (Block M then Block A).
3
Calculate the internal permutations within each block.
Internal arrangements for appetizers = 3!=63! = 6, and internal arrangements for main courses = 3!=63! = 6.
Each category consists of 3 distinct dishes that can be permuted among themselves.
4
Apply the Fundamental Counting Principle to find total linear arrangements.
Total arrangements = 2!×3!×3!=2×6×6=722! \times 3! \times 3! = 2 \times 6 \times 6 = 72.
The decision steps are independent, so their possible outcomes are multiplied.

Key Concept

Permutations with Block Restrictions (Grouping Method)

Practice More

Try solving a problem where only one specific subset of items must remain together while the rest can be arranged freely.

Alternative Method

Alternatively, place the first appetizer in slot 1 (3 choices). The remaining appetizers must take slots 2 and 3 (2 choices and 1 choice). Then slots 4, 5, 6 must be filled by main courses (3 choices, 2 choices, 1 choice). This gives 3×2×1×3×2×1=363 \times 2 \times 1 \times 3 \times 2 \times 1 = 36 ways when appetizers are placed first. Symmetrically, placing main courses in slots 1 to 3 gives another 36 ways, totaling 36+36=7236 + 36 = 72 ways.
Estimated Time:1m 30s
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Topics

Permutations and Linear Arrangements

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