A committee of three people is to be chosen from four married couples. What is the number of different committees...

1. Translate the problem requirements: We have 8 people (4 married couples) and need to select 3 people for a committee, but no married couple can both be on the committee. We need to count all valid ways to form such committees.
2. Identify the constraint pattern: Recognize that the restriction means at most one person from each married couple can be selected, so we're essentially choosing people while respecting couple boundaries.
3. Use complementary counting strategy: Calculate total possible committees of 3 people from 8, then subtract the invalid committees (those containing married couples).
4. Count invalid scenarios systematically: Determine how many committees would contain exactly one married couple plus one additional person from the remaining couples.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're dealing with in everyday terms:

We have 4 married couples, which means we have 8 people total. Think of them as:

• Couple 1: Person A and Person B
• Couple 2: Person C and Person D
• Couple 3: Person E and Person F
• Couple 4: Person G and Person H

We need to pick exactly 3 people to form a committee. However, there's an important rule: if we pick one person from a married couple, we cannot pick their spouse.

So we're looking for all the different ways to choose 3 people from these 8 people, while making sure no married couple appears together on the committee.

Process Skill: TRANSLATE - Converting the word problem into clear mathematical requirements

2. Identify the constraint pattern

Let's think about this restriction more clearly. The key insight is that from each married couple, we can pick at most one person.

This means:

• We could pick 3 people from 3 different couples (leaving 1 couple with no one selected)
• We cannot pick 2 people from the same couple plus 1 from another
• We cannot pick all 3 people from fewer than 3 couples

So essentially, we're choosing 3 couples first, then picking exactly one person from each of those 3 couples.

3. Use complementary counting strategy

Now let's count this step by step using a clear, logical approach:

Method: Count total possibilities, then subtract the invalid ones

Step 1: Total ways to pick any 3 people from 8 people

If there were no restrictions, we could pick any 3 people from 8 people in \(\mathrm{C(8,3)}\) ways.

\(\mathrm{C(8,3)} = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56\) ways

Step 2: Count the invalid committees (those with married couples)

Invalid committees are those that contain at least one married couple.

Process Skill: APPLY CONSTRAINTS - Systematically identifying what makes a committee invalid

4. Count invalid scenarios systematically

An invalid committee contains exactly one married couple plus one additional person.

Step 1: Choose which married couple will be on the committee

We have 4 married couples, so there are 4 ways to choose which couple.

Step 2: Choose the third person

Once we've selected a married couple (2 people), we need 1 more person for our committee of 3.

This third person must come from the remaining 6 people (since we can't pick from the same couple).

So there are 6 ways to choose the third person.

Step 3: Calculate total invalid committees

Total invalid committees = \(4 \times 6 = 24\) invalid committees

Step 4: Apply complementary counting

Valid committees = Total committees - Invalid committees

Valid committees = \(56 - 24 = 32\)

Process Skill: CONSIDER ALL CASES - Ensuring we account for all possible invalid scenarios

Final Answer

The number of different valid committees is 32.

This matches answer choice E. 32.

Verification: Let's double-check with an alternative approach. We need to choose 3 couples from 4 couples, then pick 1 person from each chosen couple:

• Ways to choose 3 couples from 4: \(\mathrm{C(4,3)} = 4\)
• Ways to pick 1 person from each of the 3 chosen couples: \(2 \times 2 \times 2 = 8\)
• Total valid committees: \(4 \times 8 = 32\) ✓

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint

Students often misinterpret "two people who are married to each other cannot both serve" as meaning "no married people can serve at all." This leads them to think they need to choose 3 people from some subset rather than understanding that they can choose married people, just not both spouses from the same couple.

2. Failing to recognize the two-step nature of the problem

Many students jump straight into complex case-by-case analysis without realizing this is essentially: (1) choose 3 couples from 4 couples, then (2) choose 1 person from each selected couple. This leads to overly complicated approaches and missed solutions.

3. Not considering complementary counting

Students often try to directly count valid committees through complex case work instead of recognizing that it's easier to count total possibilities minus invalid ones. This makes the problem much harder than necessary.

Errors while executing the approach

1. Arithmetic errors in combinations

When calculating \(\mathrm{C(8,3)} = 56\), students frequently make errors like forgetting to divide by 3! or making multiplication mistakes when computing \(\frac{8 \times 7 \times 6}{3 \times 2 \times 1}\). These basic computational errors lead to wrong totals.

2. Incorrect counting of invalid committees

Students often miscount the number of ways to choose the third person after selecting a married couple. They might count all 8 people instead of the remaining 6, or forget to account for the constraint that the third person cannot be from the same couple.

3. Double-counting or missing cases

When using direct counting methods, students frequently either count the same committee multiple times (like counting couple A + person E as different from couple A + person F when they meant to count A-husband + E) or miss entire categories of valid committees.

Errors while selecting the answer

1. Calculation verification errors

Students may arrive at 32 through their primary method but make errors when double-checking with the alternative approach (\(4 \times 2^3\)), leading them to doubt their correct answer and change it to a wrong choice.

2. Confusing the constraint direction

Even after solving correctly, some students second-guess themselves about whether they counted valid or invalid committees, potentially leading them to select 24 (the number of invalid committees) instead of 32 (the number of valid committees).