A music teacher is preparing for a recital in which 3 pairs of students, pair P, pair Q, and pair...

1. Translate the problem requirements: We need to count the total number of ways to order student introductions, which involves two separate decisions: (1) deciding the order of the 3 pairs (P, Q, R), and (2) deciding the order of the 2 students within each pair for their introduction.
2. Count the ways to order the pairs: Determine how many different sequences are possible for the 3 pairs P, Q, and R to perform.
3. Count the ways to order students within each pair: For each pair, determine how many ways the 2 students can be introduced in order.
4. Apply the multiplication principle: Combine the ordering decisions to find the total number of different introduction sequences possible.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we need to count here. We have 3 pairs of students (P, Q, and R), and we need to find the total number of ways to order student introductions.

This involves two separate decisions:

• First decision: In what order will the 3 pairs perform? (P first, then Q, then R? Or maybe R first, then P, then Q? etc.)
• Second decision: Within each pair, which student gets introduced first?

Think of it like this: imagine you're the announcer at the recital. You need to decide which pair to call up first, second, and third. Then, for each pair, you need to decide which of the two students in that pair to introduce first.

Process Skill: TRANSLATE - Converting the problem setup into clear counting decisions

2. Count the ways to order the pairs

Let's start with the simpler question: in how many ways can we arrange the 3 pairs P, Q, and R?

Imagine we have 3 slots on stage: [First] [Second] [Third]

• For the first slot, we can choose any of the 3 pairs: P, Q, or R
• For the second slot, we can choose any of the remaining 2 pairs
• For the third slot, we must use the last remaining pair

So we have: \(\mathrm{3} \times \mathrm{2} \times \mathrm{1} = \mathrm{6}\) ways

We can list them out to verify: PQR, PRQ, QPR, QRP, RPQ, RQP

Technically, this is \(\mathrm{3!} = \mathrm{6}\) arrangements of the pairs.

3. Count the ways to order students within each pair

Now let's think about each individual pair. Each pair has 2 students, and we need to decide the order for introducing them.

For any pair with students A and B:

• We can introduce A first, then B
• Or we can introduce B first, then A

So for each pair, there are 2 ways to order the student introductions.

Since we have 3 pairs total, and each pair has 2 possible introduction orders:

• Pair P: 2 ways to order introductions
• Pair Q: 2 ways to order introductions
• Pair R: 2 ways to order introductions

Total ways to order students within all pairs = \(\mathrm{2} \times \mathrm{2} \times \mathrm{2} = \mathrm{8}\) ways

Technically, this is \(\mathrm{2}^3 = \mathrm{8}\) combinations of introduction orders.

4. Apply the multiplication principle

Now we combine our two types of decisions using the multiplication principle.

We found:

• 6 ways to order the pairs
• 8 ways to order student introductions within the pairs

Since these decisions are independent (the order of pairs doesn't affect how we introduce students within each pair), we multiply:

Total number of different orderings = (Ways to order pairs) × (Ways to order introductions within pairs)
Total = \(\mathrm{6} \times \mathrm{8} = \mathrm{48}\)

Let's verify this makes sense: for each of the 6 possible pair orders (like PQR or QPR), we can have any of the 8 possible combinations of student introduction orders within those pairs.

Final Answer

The total number of different orderings of student introductions is 48.

This matches answer choice C.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the problem structure as a single permutation

Students often see "ordering of student introductions" and think they need to arrange all 6 students in a line, leading them to calculate \(\mathrm{6!} = \mathrm{720}\). They miss that this is actually a two-stage decision process: first ordering the pairs, then ordering students within each pair.

2. Confusing "order of performance" with "order of introductions"

Some students might think the question is asking only about the order in which pairs perform (which would be just \(\mathrm{3!} = \mathrm{6}\)), not recognizing that within each pair's performance, there's also a decision about which student gets introduced first.

3. Not recognizing this as a multiplication principle problem

Students may struggle to identify that there are two independent decisions being made simultaneously. They might try to add the possibilities instead of multiplying them, or attempt to use more complex counting methods when the multiplication principle is the key insight.

Errors while executing the approach

1. Miscounting the number of ways to order students within pairs

When calculating \(\mathrm{2}^3\) for the internal pair orderings, students might incorrectly calculate this as \(\mathrm{2} \times \mathrm{3} = \mathrm{6}\) instead of \(\mathrm{2} \times \mathrm{2} \times \mathrm{2} = \mathrm{8}\), confusing the number of pairs (3) with the number of arrangements per pair (2).

2. Arithmetic errors in the final multiplication

Even with the correct setup of \(\mathrm{6} \times \mathrm{8}\), students might make simple multiplication errors, potentially calculating \(\mathrm{6} \times \mathrm{8}\) as 42 or 54, especially under time pressure during the exam.

Errors while selecting the answer

No likely faltering points