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Difficulty: MediumPermutations and Combinations

A project manager must schedule 5 distinct client presentations—for clients A, B, C, D, and E—on 5 consecutive days from Monday through Friday, with exactly one presentation per day. If the presentations for client A and client B cannot be scheduled on consecutive days, how many different presentation schedules are possible?

Answer: 72 schedules

Answer

The total number of valid presentation schedules is 72.
The correct result is found by subtracting the number of restricted arrangements (where presentation A and presentation B are scheduled on consecutive days) from the total number of unrestricted arrangements of 5 presentations. The total unrestricted arrangements equal 5!=1205! = 120. Treating A and B as a single block leaves 4 items to arrange in 4!=244! = 24 ways, with 2!=22! = 2 internal orderings for A and B, yielding 24×2=4824 \times 2 = 48 consecutive schedules. Subtracting 48 from 120 results in 72 valid schedules.

Step-by-Step Solution

1
Calculate total unrestricted linear arrangements of the 5 presentations.
5!=1205! = 120
Without restrictions, 5 distinct items can be arranged in 5 distinct positions in 5!5! ways.
2
Calculate the number of invalid arrangements where presentations A and B are on consecutive days.
2!×4!=482! \times 4! = 48
Grouping A and B into a single unit results in 4 items to arrange (4!=244! = 24), and A and B can swap positions inside the block in 2!=22! = 2 ways.
3
Subtract the invalid arrangements from total arrangements.
120−48=72120 - 48 = 72
Complementary counting gives the total number of arrangements where A and B are not on consecutive days.

Key Concept

Permutations with Adjacency Restrictions (Complementary Counting)
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Topics

Permutations and Combinations

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