If a committee of 3 men and 3 women is to be selected from a group of 7 men and...

1. Translate the problem requirements: We need to form a committee with exactly 3 men and exactly 3 women. From 7 available men, we must choose 3. From 7 available women, we must choose 3. The question asks for the total number of different possible committees.
2. Recognize independent selection processes: Since selecting men and women are separate, independent processes, we can calculate each selection separately and then multiply the results.
3. Calculate combinations for each group: Use the combination formula to find how many ways we can select 3 men from 7 men, and 3 women from 7 women.
4. Apply the multiplication principle: Multiply the number of ways to select men by the number of ways to select women to get the total number of different committees.
5. Match with answer choices: Verify our calculated result corresponds to the given answer format.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking for in everyday language. We have a group of 14 people total: 7 men and 7 women. We want to form a committee that has exactly 3 men and exactly 3 women. The question asks: how many different ways can we form such a committee?

Think of it like this: imagine you're organizing a small planning committee for an event. You specifically want equal representation, so you need exactly 3 men and exactly 3 women on your 6-person committee. The key insight is that we need to count all the possible combinations of people we could choose.

Process Skill: TRANSLATE - Converting the problem language into a clear understanding of what we need to find

2. Recognize independent selection processes

Here's an important insight: choosing the men for our committee is completely separate from choosing the women. Think about it this way - once we decide which 3 men will be on the committee, that decision doesn't affect which 3 women we can choose. These are independent choices.

For example, if we choose men named Alex, Bob, and Carl, we could still pair them with any combination of 3 women from our group of 7 women. Similarly, if we choose women named Diana, Emma, and Fiona, we could pair them with any combination of 3 men from our group of 7 men.

This independence is crucial because it means we can calculate the number of ways to select men separately from the number of ways to select women, then combine these results.

3. Calculate combinations for each group

Now let's figure out how many ways we can make each selection.

For the men: We need to choose 3 men from a group of 7 men. This is a 'combination' problem because the order doesn't matter - a committee with Alex, Bob, and Carl is the same committee whether we picked Alex first or Carl first.

The number of ways to choose 3 items from 7 items is written as \(\mathrm{C(7,3)}\) or "7 choose 3".
Using the combination formula: \(\mathrm{C(7,3) = 7!/(3! \times 4!) = (7 \times 6 \times 5)/(3 \times 2 \times 1) = 210/6 = 35}\)

For the women: By the exact same reasoning, we need to choose 3 women from 7 women.
\(\mathrm{C(7,3) = 35}\) (same calculation as above)

So we have 35 ways to choose the men and 35 ways to choose the women.

4. Apply the multiplication principle

Since selecting men and selecting women are independent processes, we multiply the number of possibilities for each.

Think of it this way: for each of the 35 ways to choose men, we can pair it with any of the 35 ways to choose women. This gives us:

Total number of different committees = \(\mathrm{35 \times 35 = 35^2}\)

This makes intuitive sense - we're creating all possible combinations by pairing every possible group of men with every possible group of women.

5. Match with answer choices

Looking at our answer choices:

1. 36
2. 48
3. \(\mathrm{24^2}\)
4. \(\mathrm{35^2}\)
5. \(\mathrm{(7!/3!)^2}\)

Our calculated result of \(\mathrm{35^2}\) matches exactly with choice D.

We can also verify that choice E gives the same result:
\(\mathrm{(7!/3!)^2 = (7 \times 6 \times 5 \times 4!)/(3! \times 4!) \times (7 \times 6 \times 5 \times 4!)/(3! \times 4!) = (7 \times 6 \times 5/3!)^2 = (210/6)^2 = 35^2}\)

Final Answer

The answer is D. \(\mathrm{35^2}\)

We found that there are 35 ways to choose 3 men from 7 men, and 35 ways to choose 3 women from 7 women. Since these selections are independent, we multiply to get \(\mathrm{35^2 = 1,225}\) total different possible committees.

Common Faltering Points

Errors while devising the approach

1. Treating this as a permutation problem instead of a combination problem
Students often get confused about when order matters. Since we're forming a committee, the order in which we select people doesn't matter - choosing Alex then Bob then Carl creates the same committee as choosing Carl then Bob then Alex. However, students might mistakenly think they need to use permutation formulas (like \(\mathrm{P(7,3)}\)) instead of combination formulas (\(\mathrm{C(7,3)}\)), leading to overcounting.

2. Attempting to select all 6 people simultaneously from 14 people without considering gender constraints
A common mistake is to think "I need to choose 6 people from 14 total people" and calculate \(\mathrm{C(14,6)}\). This completely ignores the crucial constraint that we need exactly 3 men AND exactly 3 women. This approach would include committees with 4 men and 2 women, or 5 women and 1 man, etc., which violates the problem requirements.

3. Failing to recognize that selecting men and women are independent processes
Students might struggle to see that choosing 3 men from 7 men is completely separate from choosing 3 women from 7 women. They might try to devise overly complex approaches instead of recognizing that these two selection processes can be calculated separately and then multiplied together.

Errors while executing the approach

1. Making arithmetic errors in combination calculations
When calculating \(\mathrm{C(7,3) = 7!/(3! \times 4!)}\), students often make computational mistakes. For example, they might incorrectly calculate \(\mathrm{7 \times 6 \times 5 = 200}\) instead of 210, or forget to divide by \(\mathrm{3! = 6}\), or make errors in the division step (\(\mathrm{210 \div 6}\)). These arithmetic slips lead to wrong intermediate results.

2. Forgetting to apply the multiplication principle
Even after correctly calculating that there are 35 ways to choose men and 35 ways to choose women, students sometimes forget to multiply these together. They might just add them (\(\mathrm{35 + 35 = 70}\)) or only report one of the values (35), missing the final crucial step of \(\mathrm{35 \times 35 = 35^2}\).

Errors while selecting the answer

1. Selecting the numerical value instead of the expression form
Students might correctly calculate \(\mathrm{35^2 = 1,225}\) but then look for 1,225 among the answer choices. When they don't find it listed explicitly, they might pick a random answer instead of recognizing that choice D (\(\mathrm{35^2}\)) represents exactly the same value in expression form.