If there are 4 pairs of twins, and a committee will be formed with 3 members. In how many ways...

1. Translate the problem requirements: We have 8 people (4 pairs of twins) and need to form a 3-person committee where no two siblings are selected together
2. Identify the constraint structure: Since no siblings can be together, we can select at most one person from each twin pair
3. Apply systematic selection logic: Choose 3 different twin pairs first, then select one person from each chosen pair
4. Calculate using basic counting principles: Multiply the ways to choose pairs by the ways to choose people from selected pairs

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have and what we need to find.

We have 4 pairs of twins, which means we have 8 people total. Let's call them:

• Twin Pair 1: Person A1 and Person A2
• Twin Pair 2: Person B1 and Person B2
• Twin Pair 3: Person C1 and Person C2
• Twin Pair 4: Person D1 and Person D2

We want to form a committee of 3 people, but here's the key constraint: no siblings can be on the committee together. This means we cannot pick both A1 and A2, or both B1 and B2, etc.

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

2. Identify the constraint structure

Since no siblings can be together, let's think about what this means for our selection:

If we pick someone from Twin Pair A (either A1 or A2), we cannot pick anyone else from Twin Pair A. The same logic applies to all other twin pairs.

Since we need 3 people for our committee and we can pick at most one person from each twin pair, we need to:

• Choose exactly 3 different twin pairs (out of our 4 available pairs)
• From each of these 3 chosen pairs, select exactly 1 person

This ensures no siblings end up on the same committee.

Process Skill: APPLY CONSTRAINTS - Recognizing how the "no siblings" rule shapes our selection strategy

3. Apply systematic selection logic

Now let's break this down into steps:

Step 1: Choose which 3 twin pairs we'll select people from
We have 4 twin pairs and need to choose 3 of them. Let's think of some examples:

• We could choose pairs A, B, and C
• We could choose pairs A, B, and D
• We could choose pairs A, C, and D
• We could choose pairs B, C, and D

How many ways can we choose 3 pairs from 4? This is simply 4 choose 3, which equals 4.

Step 2: For each chosen set of 3 pairs, decide which person to pick from each pair
Once we've decided on our 3 pairs, we need to pick one person from each pair:

• From the first chosen pair: 2 options (either twin)
• From the second chosen pair: 2 options (either twin)
• From the third chosen pair: 2 options (either twin)

So for any set of 3 pairs, we have \(\mathrm{2 × 2 × 2 = 8}\) ways to form our committee.

4. Calculate using basic counting principles

Now we multiply our results from the two steps:

• Number of ways to choose 3 pairs from 4 pairs = 4
• Number of ways to choose people from the selected 3 pairs = 8

Total number of ways = \(\mathrm{4 × 8 = 32}\)

Let's verify with a concrete example:
If we choose pairs A, B, and C:

• A1-B1-C1, A1-B1-C2, A1-B2-C1, A1-B2-C2
• A2-B1-C1, A2-B1-C2, A2-B2-C1, A2-B2-C2

That's 8 committees from just one choice of 3 pairs. Since we have 4 different ways to choose our 3 pairs, we get \(\mathrm{4 × 8 = 32}\) total committees.

Final Answer

The number of ways to form a 3-member committee from 4 pairs of twins with no siblings together is 32.

This matches answer choice A.

Common Faltering Points

Errors while devising the approach

Students often misinterpret what "no siblings in a group" means. They might think it means "no twins at all" rather than "no two people from the same twin pair." This leads them to believe they need to select 3 people from some subset rather than understanding that they can select one person from each of three different twin pairs.

Many students try to solve this as a straightforward combination problem (selecting 3 from 8 people with restrictions) rather than recognizing that this is actually a two-step process: first choosing which twin pairs to include, then choosing which twin from each selected pair. This approach becomes very complicated and error-prone.

Students might not realize that since we need 3 committee members and can't have siblings together, we must select exactly 3 different twin pairs and exactly 1 person from each pair. They may try approaches that allow for selecting 0 or 2+ people from certain pairs.

Errors while executing the approach

When calculating the number of ways to choose 3 pairs from 4 pairs, students might incorrectly compute \(\mathrm{C(4,3)}\). Some may calculate it as \(\mathrm{3! = 6}\) instead of the correct value of 4, or confuse it with other combination formulas.

After correctly identifying that they need to choose 3 pairs from 4, students might forget that for each selected pair, they have 2 choices (which twin to pick). They may calculate only \(\mathrm{C(4,3) = 4}\) and miss the multiplication by \(\mathrm{2^3 = 8}\).

Even if students identify both steps correctly, they might make errors in applying the multiplication principle. For example, they might add \(\mathrm{4 + 8 = 12}\) instead of multiplying \(\mathrm{4 × 8 = 32}\), or they might incorrectly calculate \(\mathrm{4 × 2 × 3 = 24}\) instead of \(\mathrm{4 × 2^3 = 32}\).

Errors while selecting the answer

No likely faltering points - the calculation directly leads to 32, which clearly corresponds to answer choice A. The final step is straightforward once the computation is complete.