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Difficulty: HardCombinations and Group Selections

A corporate committee needs to form a 5-member project team selected from a department consisting of 4 data analysts, 5 software developers, and 4 cybersecurity experts. The team must include exactly 1 cybersecurity expert, at least 1 data analyst, and at least 1 software developer. How many different 5-member teams can be formed under these conditions?

  1. A
    124
  2. B
    240
  3. 480Answer
  4. D
    504
  5. E
    1,120

Answer

480
To form a valid 5-member team, we first choose 1 cybersecurity expert from 4 available options in (41)=4\binom{4}{1} = 4 ways. The remaining 4 members must be selected from 4 data analysts and 5 software developers such that at least 1 of each role is included. The three mutually exclusive ways to fill these 4 spots are: 1 data analyst and 3 developers (40 ways), 2 data analysts and 2 developers (60 ways), or 3 data analysts and 1 developer (20 ways). Summing these gives 120 valid combinations for the remaining members. Multiplying by the 4 choices for the cybersecurity expert yields 4×120=4804 \times 120 = 480 total distinct teams.

Step-by-Step Solution

1
Determine the number of ways to select the required cybersecurity expert
4 ways
Selecting exactly 1 cybersecurity expert from a total of 4 available experts is given by the combination (41)=4\binom{4}{1} = 4.
2
Identify the possible valid distributions for the remaining 4 team members
Three valid cases: (1 DA, 3 SD), (2 DA, 2 SD), and (3 DA, 1 SD)
The team requires 5 members in total. With 1 cybersecurity expert selected, 4 positions remain. Since the team must contain at least 1 data analyst (DA) and at least 1 software developer (SD), we list all valid splits of 4 members.
3
Calculate the combinations for each valid distribution of data analysts and software developers
Case 1: 40 ways; Case 2: 60 ways; Case 3: 20 ways
Case 1 (1 DA, 3 SD): (41)×(53)=4×10=40\binom{4}{1} \times \binom{5}{3} = 4 \times 10 = 40.
Case 2 (2 DA, 2 SD): (42)×(52)=6×10=60\binom{4}{2} \times \binom{5}{2} = 6 \times 10 = 60.
Case 3 (3 DA, 1 SD): (43)×(51)=4×5=20\binom{4}{3} \times \binom{5}{1} = 4 \times 5 = 20.
4
Sum the valid remaining member distributions and apply the Fundamental Counting Principle
480 total teams
Total remaining combinations = 40+60+20=12040 + 60 + 20 = 120. Total team combinations = (41)×120=4×120=480\binom{4}{1} \times 120 = 4 \times 120 = 480.

Key Concept

Group selections with multiple category constraints and the Fundamental Counting Principle
Estimated Time:2m 30s
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Topics

Combinations and Group Selections

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