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Difficulty: MediumPermutations and Linear Arrangements

A corporate board of 7 members—consisting of 4 senior executives (including the CEO and the COO) and 3 junior associates—is to be seated in a single row of 7 chairs for a press conference. If the 3 junior associates must sit in adjacent chairs, and the CEO and the COO cannot sit next to each other, in how many different linear arrangements can the 7 members be seated?

  1. A
    288
  2. 432Answer
  3. C
    576
  4. D
    720
  5. E
    1,440

Answer

432
To satisfy the condition that the 3 junior associates sit together, we treat them as 1 block with 3!=63! = 6 internal orderings. Combining this block with the 4 senior executives yields 5 entities, which can be arranged in 5!=1205! = 120 ways, giving 120×6=720120 \times 6 = 720 total arrangements with the juniors seated together. To enforce that the CEO and COO cannot sit together, we subtract the arrangements where they do sit together: treating the CEO and COO as a block gives 4 entities to arrange (4!=244! = 24), with 2!=22! = 2 ways to arrange CEO and COO, and 3!=63! = 6 ways for the junior block, totaling 24×2×6=28824 \times 2 \times 6 = 288 restricted cases. Subtracting 288 from 720 gives 432 valid arrangements.

Step-by-Step Solution

1
Group the 3 junior associates into a single block.
The 3 junior associates can be arranged internally within their block in 3!=63! = 6 ways. Treating this block as 1 single element along with the 4 senior executives gives a total of 5 items to arrange.
The condition specifies that all 3 junior associates must sit in adjacent chairs.
2
Calculate total arrangements where junior associates sit together without CEO/COO restrictions.
The 5 items (1 block + 4 senior executives) can be arranged in 5!=1205! = 120 ways. Including internal block arrangements gives 120×6=720120 \times 6 = 720 ways.
Applying the Fundamental Counting Principle to the 5 units and the 3 internal positions.
3
Calculate unwanted arrangements where the CEO and COO sit next to each other (with junior associates together).
Group the CEO and COO into a second block with 2!=22! = 2 internal arrangements. Now there are 4 items to arrange (Junior block, CEO-COO block, and 2 other senior executives). Total unwanted arrangements: 4!×3!×2!=24×6×2=2884! \times 3! \times 2! = 24 \times 6 \times 2 = 288 ways.
To find valid non-adjacent arrangements, subtract adjacent CEO-COO arrangements from total junior-grouped arrangements.
4
Subtract the unwanted arrangements from the total grouped arrangements.
720−288=432720 - 288 = 432 valid seating arrangements.
Complementary counting provides a direct solution.

Key Concept

Linear Permutations with Grouping and Complementary Non-Adjacency Constraints
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Topics

Permutations and Linear Arrangements

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