A software development team consists of 4 front-end developers, 5 back-end developers, and… — ACT

A software development team consists of $4$ front-end developers, $5$ back-end developers, and $3$ quality assurance (QA) engineers. The project manager needs to select a $3$ -member subcommittee to present a new product feature. If the subcommittee must contain at least $1$ front-end developer and at least $1$ back-end developer, how many different $3$ -member subcommittees can be formed?

Answer: 130 subcommittees

Answer

The total number of valid 3-member subcommittees that can be formed is 130.

To form a 3-member committee with at least 1 front-end developer and at least 1 back-end developer from 4 front-end, 5 back-end, and 3 QA engineers, we break the problem into three mutually exclusive valid scenarios: selecting 1 member from each role gives

4 \times 5 \times 3 = 60

ways; selecting 2 front-end and 1 back-end developer gives

C(4,2) \times C(5,1) = 6 \times 5 = 30

ways; and selecting 1 front-end and 2 back-end developers gives

C(4,1) \times C(5,2) = 4 \times 10 = 40

ways. Summing these possibilities gives

60 + 30 + 40 = 130

. Alternatively, subtracting invalid committees (those with no front-end developers:

C(8,3) = 56

, no back-end developers:

C(7,3) = 35

, minus double-counted 3 QA engineers:

1

) from total 3-member committees (

C(12,3) = 220

) yields

220 - (56 + 35 - 1) = 130

Step-by-Step Solution

Determine the required subcommittee size and role counts

The team has 4 front-end developers, 5 back-end developers, and 3 QA engineers (12 members total). A subcommittee of size 3 is required.

Establishing the total population and category counts is necessary before computing combination constraints.

Enumerate the mutually exclusive cases that satisfy all constraints

Case 1: (1 front-end, 1 back-end, 1 QA)
Case 2: (2 front-end, 1 back-end, 0 QA)
Case 3: (1 front-end, 2 back-end, 0 QA)

The committee requires at least 1 front-end and at least 1 back-end member in a 3-person group.

Compute combinations for each valid case using the combination formula

C(n, k) = \frac{n!}{k!(n-k)!}

Case 1:

C(4,1) \times C(5,1) \times C(3,1) = 4 \times 5 \times 3 = 60

Case 2:

C(4,2) \times C(5,1) \times C(3,0) = 6 \times 5 \times 1 = 30

Case 3:

C(4,1) \times C(5,2) \times C(3,0) = 4 \times 10 \times 1 = 40

Order of selection does not matter when forming a committee, so combinations are used.

Add the counts from all mutually exclusive cases

Total valid subcommittees =

60 + 30 + 40 = 130

According to the addition rule of counting, the total number of outcomes across disjoint cases is the sum of their individual counts.

Key Concept

Counting combinations across multiple groups with specific distribution constraints

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