From a group of 21 astronauts that includes 12 people with previous experience in space flight, a 3-person crew is...

1. Translate the problem requirements: We need to clarify what "exactly one person with experience" means - we're selecting 1 experienced person from 12 available, and 2 inexperienced people from the remaining 9 people (since 21 total - 12 experienced = 9 inexperienced).
2. Identify the selection components: Break down this combination problem into two separate, independent selection tasks that must both occur.
3. Calculate each selection separately: Find the number of ways to choose 1 from the experienced group, then find the number of ways to choose 2 from the inexperienced group.
4. Apply the multiplication principle: Since both selections must happen together to form our crew, multiply the results from each independent selection.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're looking for in plain English. We have 21 astronauts total, and we know that 12 of them have previous space experience. This means the remaining astronauts (\(21 - 12 = 9\)) have no previous space experience.

We need to form a 3-person crew where exactly one person has experience. Think of it like this: if I'm building this crew, I need to pick:

• Exactly 1 person from the 12 experienced astronauts
• Exactly 2 people from the 9 inexperienced astronauts

This gives me my crew of 3 people total, with exactly 1 experienced member.

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

2. Identify the selection components

Now I can see this problem has two separate selection tasks that I need to do independently:

Task A: Choose 1 experienced astronaut from the 12 available
Task B: Choose 2 inexperienced astronauts from the 9 available

These are two completely separate choices. My selection of which experienced astronaut to pick doesn't affect which inexperienced astronauts I can choose, and vice versa.

3. Calculate each selection separately

Let's work out each selection:

Task A: Choosing 1 from 12 experienced astronauts
This is straightforward - if I need to pick 1 person from 12 people, there are simply 12 different ways to do this. I could pick astronaut #1, or astronaut #2, or astronaut #3, and so on up to astronaut #12.
Number of ways = 12

Task B: Choosing 2 from 9 inexperienced astronauts
This is a bit more involved. Let me think about it step by step. If I'm choosing 2 people from 9, I need to count combinations (not arrangements, since the order doesn't matter in a crew).

I can use the combination formula, but let me first explain it intuitively:

• For my first choice, I have 9 options
• For my second choice, I have 8 remaining options
• This gives me \(9 \times 8 = 72\) ways
• But this counts each pair twice (choosing person A then person B is the same crew as choosing person B then person A)
• So I divide by 2: \(72 \div 2 = 36\) ways

Technically, this is \(\mathrm{C}(9,2) = \frac{9!}{2! \times 7!} = \frac{9 \times 8}{2 \times 1} = 36\)

4. Apply the multiplication principle

Now I need to combine these selections. Since I must do BOTH Task A AND Task B to form my complete crew, I multiply the number of ways:

Total number of different crews = (Ways to choose 1 experienced) × (Ways to choose 2 inexperienced)
Total = \(12 \times 36 = 432\)

Let me verify this makes sense: each of the 12 experienced astronauts could be paired with any of the 36 possible pairs of inexperienced astronauts, giving us \(12 \times 36 = 432\) unique crew combinations.

Final Answer

The number of different crews possible is 432.

Looking at our answer choices, this matches choice A. 432.

Answer: A

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting "exactly one" as "at least one"
Students often misread the constraint and think they need crews where one OR MORE people have experience, rather than EXACTLY one. This leads them to calculate multiple scenarios (1 experienced + 2 inexperienced, 2 experienced + 1 inexperienced, etc.) instead of the single required scenario.

Faltering Point 2: Failing to recognize the two separate selection tasks
Some students try to approach this as a single combination problem C(21,3) and then attempt complex adjustments. They don't break down the problem into the two independent selection components: choosing experienced astronauts and choosing inexperienced astronauts separately.

Faltering Point 3: Incorrectly calculating the number of inexperienced astronauts
Students may forget to subtract and assume there are 12 inexperienced astronauts (same as experienced), or make arithmetic errors when calculating \(21 - 12 = 9\). This error in the basic setup leads to wrong numbers in all subsequent calculations.

Errors while executing the approach

Faltering Point 1: Confusing combinations with permutations
When calculating C(9,2), students might calculate \(9 \times 8 = 72\) but forget to divide by 2!, thinking that order matters in crew selection. This gives them 72 instead of the correct 36, leading to a final answer of \(12 \times 72 = 864\).

Faltering Point 2: Arithmetic errors in combination calculations
Students may make computational mistakes when calculating \(\mathrm{C}(9,2) = \frac{9!}{2! \times 7!}\), either in the factorial arithmetic or in the division step, leading to incorrect intermediate results.

Errors while selecting the answer

Faltering Point 1: Choosing 864 due to the permutation error
If students made the permutation/combination error in execution (getting 72 instead of 36), they would calculate \(12 \times 72 = 864\) and select answer choice C instead of the correct answer A.