In a certain group of 10 members, 4 members teach only French and the rest teach only Spanish or German....

1. Translate the problem requirements: We have 10 members total: 4 teach only French, 6 teach only Spanish or German. We need to form a 3-member committee that must include at least 1 French teacher. We need to count all possible different committees.
2. Identify the constraint strategy: Rather than directly counting committees with "at least 1 French teacher," we'll use the complement approach - count all possible committees and subtract those with no French teachers.
3. Calculate total possible committees: Find the number of ways to choose any 3 members from the 10 total members without restrictions.
4. Calculate restricted committees to subtract: Find the number of ways to choose 3 members from only the 6 non-French teachers (Spanish/German only).
5. Apply complement principle: Subtract the restricted count from the total count to get committees with at least 1 French teacher.

Execution of Strategic Approach

1. Translate the problem requirements

Let's carefully break down what we're told in plain English:

• We have a group of 10 members total
• 4 members teach only French
• The remaining members (10 - 4 = 6 members) teach only Spanish or German
• We need to form a committee of exactly 3 members
• The constraint is: the committee MUST have at least 1 member who teaches French
• We want to count how many different committees are possible

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

2. Identify the constraint strategy

When we see "at least 1" in a counting problem, we have two approaches:

Direct approach: Count committees with exactly 1 French teacher + committees with exactly 2 French teachers + committees with exactly 3 French teachers

Complement approach: Count ALL possible committees, then subtract the committees that violate our constraint (committees with 0 French teachers)

The complement approach is usually simpler because instead of adding multiple cases, we just do one subtraction. Let's use this approach:

Committees with at least 1 French teacher = All possible committees - Committees with 0 French teachers

Process Skill: CONSIDER ALL CASES - Choosing the most efficient counting strategy

3. Calculate total possible committees

First, let's find how many ways we can choose any 3 members from all 10 members, with no restrictions.

Think of it this way: we're selecting 3 people from 10 people. This is a combination problem because the order doesn't matter (a committee with members A, B, C is the same as a committee with members B, A, C).

Using the combination formula: \(\mathrm{C(10,3)} = \frac{10!}{3! \times 7!}\)

Let's calculate step by step:
• \(\mathrm{C(10,3)} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}\)
• \(\mathrm{C(10,3)} = \frac{720}{6} = 120\)

So there are 120 total ways to form any 3-member committee from 10 people.

4. Calculate restricted committees to subtract

Now we need to find how many committees have 0 French teachers (which means all 3 members come from the 6 non-French teachers).

We're choosing 3 members from the 6 Spanish/German teachers:

\(\mathrm{C(6,3)} = \frac{6!}{3! \times 3!}\)
• \(\mathrm{C(6,3)} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1}\)
• \(\mathrm{C(6,3)} = \frac{120}{6} = 20\)

So there are 20 ways to form a committee with no French teachers.

5. Apply complement principle

Now we can find the answer using our complement approach:

Committees with at least 1 French teacher = Total committees - Committees with 0 French teachers

Committees with at least 1 French teacher = \(120 - 20 = 100\)

Process Skill: APPLY CONSTRAINTS - Using complement counting to handle the "at least" condition efficiently

Final Answer

There are 100 different committees that can be chosen.

Looking at our answer choices:

1. 40
2. 50
3. 64
4. 80
5. 100

Our answer of 100 matches choice E.

Verification: Let's double-check by using the direct method:
• Committees with exactly 1 French teacher: \(\mathrm{C(4,1)} \times \mathrm{C(6,2)} = 4 \times 15 = 60\)
• Committees with exactly 2 French teachers: \(\mathrm{C(4,2)} \times \mathrm{C(6,1)} = 6 \times 6 = 36\)
• Committees with exactly 3 French teachers: \(\mathrm{C(4,3)} \times \mathrm{C(6,0)} = 4 \times 1 = 4\)
• Total: \(60 + 36 + 4 = 100\) ✓

Answer: E. 100

Common Faltering Points

Errors while devising the approach

1. Misinterpreting the constraint "at least 1 member who teaches French"
Students often confuse "at least 1" with "exactly 1" and only count committees with exactly one French teacher, missing committees with 2 or 3 French teachers. This leads to undercounting the total possibilities.

2. Not recognizing the complement approach as more efficient
Many students attempt the direct method (counting exactly 1, exactly 2, exactly 3 French teachers separately) without considering that the complement approach (total minus zero French teachers) is simpler. While both methods work, the direct approach has more steps where errors can occur.

3. Misunderstanding the group composition
Some students misread "4 members teach only French and the rest teach only Spanish or German" and incorrectly think there are separate counts for Spanish and German teachers, rather than understanding that the remaining 6 members form one group of non-French teachers.

Errors while executing the approach

1. Combination formula calculation errors
Students frequently make arithmetic mistakes when calculating C(10,3) = 120 or C(6,3) = 20, especially in the factorial expansions. Common errors include miscalculating (10×9×8)/(3×2×1) or (6×5×4)/(3×2×1).

2. Confusing combinations with permutations
Some students incorrectly use permutation formulas instead of combination formulas, thinking that the order of selection matters for committee formation. This leads to overcounting by factors of 3! = 6.

Errors while selecting the answer

1. Selecting an intermediate calculation as the final answer
Students sometimes select 120 (total committees) or 20 (committees with no French teachers) instead of the final result 100. This happens when they lose track of what the question is actually asking for after performing multiple calculations.