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If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?
Let's start by understanding what we have and what we need to find.
We have 5 married couples, which means we have 10 people total. Think of them as:
We need to select 3 people for a committee, but here's the key constraint: we cannot select both people from the same couple. So we can't have both Person A and Person B on the committee, for example.
Process Skill: TRANSLATE - Converting the problem language into a clear mathematical understanding
Now let's think about what this constraint really means in practical terms.
Since we need 3 people and we can't take both people from any couple, we must select our 3 people from 3 different couples. There's no other way to do it!
Think about it: if we tried to select people from only 2 couples, we could get at most 2 people (one from each couple). But we need 3 people, so we must use 3 different couples.
This breaks our problem into two simpler steps:
Process Skill: INFER - Drawing the non-obvious conclusion that we must select from exactly 3 different couples
Let's figure out how many ways we can choose 3 couples from our 5 available couples.
We can think of this step by step:
If order mattered, this would give us \(\mathrm{5 \times 4 \times 3 = 60}\) ways.
But the order doesn't matter! Choosing couples (1, 2, 3) is the same as choosing couples (2, 1, 3) or any other arrangement. Since we can arrange 3 couples in \(\mathrm{3! = 6}\) different ways, we need to divide by 6.
Number of ways to choose 3 couples = \(\mathrm{(5 \times 4 \times 3) \div 6 = 60 \div 6 = 10}\)
Now, for each way of choosing 3 couples, we need to figure out how many ways we can select one person from each couple.
Let's say we've chosen couples 1, 2, and 3. Now:
Since these choices are independent, we multiply: \(\mathrm{2 \times 2 \times 2 = 8}\) ways to select people from any set of 3 chosen couples.
Now we can combine our results using the multiplication principle.
We found:
Since these are sequential decisions (first choose the couples, then choose the people), we multiply:
Total number of committees = \(\mathrm{10 \times 8 = 80}\)
The number of possible committees is 80.
Looking at our answer choices, this matches choice D.
To verify: We systematically addressed the constraint by recognizing that we must select from exactly 3 different couples, calculated the combinations properly, and applied the multiplication principle correctly. Our answer of 80 committees is correct.
Students often misread "the committee does not include two people who are married to each other" as meaning we can only select unmarried people, rather than understanding that we simply cannot select both spouses from the same couple. This leads to completely incorrect problem setup.
Many students immediately jump to calculating \(\mathrm{C(10,3) = 120}\) total ways to select 3 people from 10, then try to subtract "bad" combinations. This approach becomes unnecessarily complex and error-prone compared to the systematic approach of first selecting couples, then selecting people.
Students may not realize that selecting 3 people with the given constraint necessarily means selecting from exactly 3 different couples. Without this insight, they struggle to break the problem into manageable steps and may attempt overcomplicated counting methods.
When calculating ways to choose 3 couples from 5, students correctly identify \(\mathrm{5 \times 4 \times 3 = 60}\) but forget that order doesn't matter. They fail to divide by \(\mathrm{3! = 6}\), leading to 60 instead of the correct 10 ways to choose couples.
Even with correct intermediate steps (10 ways to choose couples, 8 ways to choose people), students may make simple multiplication errors: \(\mathrm{10 \times 8}\), potentially getting 18, 88, or other incorrect results due to careless calculation.
No likely faltering points - the final step is straightforward once the calculation is complete, and the computed answer of 80 directly matches one of the given choices.