Question

Difficulty: MediumBasic Probability and Counting Methods

A conference schedule allows attendees to select 1 morning workshop out of 6 options, 1 keynote session out of 3 options, and 1 afternoon panel out of 4 options. However, 2 specific morning workshops conflict with 1 specific afternoon panel and cannot be chosen together. How many different valid 3-session schedule combinations can an attendee create?

Answer: 66 combinations

Answer

An attendee can create 66 different valid 3-session schedule combinations.
According to the Fundamental Counting Principle, the total unrestricted number of schedule combinations is 6×3×4=726 \times 3 \times 4 = 72. The restriction specifies that 2 morning workshops cannot be paired with 1 specific afternoon panel. Since there are 3 keynote speaker options available for any schedule, the number of invalid combinations is 2×3×1=62 \times 3 \times 1 = 6. Subtracting the invalid options from the total gives 726=6672 - 6 = 66 valid schedule combinations.

Step-by-Step Solution

1
Calculate total unrestricted schedule combinations
72 combinations
Multiply the choices for each session: 6×3×4=726 \times 3 \times 4 = 72.
2
Calculate the number of conflicting schedule combinations
6 invalid combinations
The 2 restricted morning workshops combined with 1 restricted afternoon panel can occur alongside any of the 3 keynote choices (2×3×1=62 \times 3 \times 1 = 6).
3
Subtract conflicting combinations from total combinations
66 valid combinations
Subtracting 6 invalid schedules from 72 total schedules leaves 66 allowable options.

Key Concept

Fundamental Counting Principle with Restrictions
Estimated Time:1m 30s
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