A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed...

1. Translate the problem requirements: We have 10 total partners (4 senior + 6 junior) and need to form groups of 3 partners where at least one member is a senior partner. "At least one senior" means 1, 2, or 3 seniors in the group.
2. Recognize the complementary counting opportunity: Instead of calculating multiple scenarios (1 senior + 2 juniors, 2 seniors + 1 junior, 3 seniors), we can use the complement: Total groups of 3 - Groups with no seniors (all juniors).
3. Calculate total possible groups: Find the number of ways to choose any 3 partners from all 10 partners.
4. Calculate groups with no seniors: Find the number of ways to choose 3 partners from only the 6 junior partners.
5. Apply complementary counting: Subtract the all-junior groups from the total groups to get groups with at least one senior.

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 a firm with:
• 4 senior partners
• 6 junior partners
• Total of 10 partners

We need to form groups of 3 partners where "at least one member is a senior partner."

What does "at least one senior" mean? It means we could have:
• 1 senior partner + 2 junior partners, OR
• 2 senior partners + 1 junior partner, OR
• 3 senior partners + 0 junior partners

So we need to count all possible 3-person groups that include 1, 2, or 3 senior partners.

Process Skill: TRANSLATE - Converting the constraint "at least one senior" into specific cases we need to count.

2. Recognize the complementary counting opportunity

Instead of calculating three separate scenarios (1 senior, 2 seniors, 3 seniors), there's a much easier approach.

Think about it this way: If we want groups with "at least one senior," that's the same as saying "all possible groups EXCEPT those with no seniors at all."

Groups with no seniors = groups with only junior partners = groups we DON'T want

So our answer will be:
Total possible 3-person groups - Groups with only juniors = Groups with at least one senior

This complementary approach turns a complex multi-case problem into just two simple calculations!

3. Calculate total possible groups

How many ways can we choose any 3 partners from all 10 partners?

Imagine we're picking 3 people one by one:
• First pick: 10 choices
• Second pick: 9 remaining choices
• Third pick: 8 remaining choices

This gives us \(\mathrm{10 \times 9 \times 8 = 720}\) ways

But wait! This counts each group multiple times because the order doesn't matter. For example, picking "Partner A, then Partner B, then Partner C" is the same group as "Partner B, then Partner C, then Partner A."

How many ways can we arrange 3 people? \(\mathrm{3 \times 2 \times 1 = 6}\) ways

So the actual number of different groups is: \(\mathrm{720 \div 6 = 120}\) total groups

Using combination notation: \(\mathrm{C(10,3) = 120}\)

4. Calculate groups with no seniors

How many ways can we choose 3 partners from only the 6 junior partners?

Using the same logic:
• First pick from juniors: 6 choices
• Second pick from remaining juniors: 5 choices
• Third pick from remaining juniors: 4 choices

This gives us \(\mathrm{6 \times 5 \times 4 = 120}\) ways

But again, order doesn't matter, so we divide by the number of ways to arrange 3 people:
\(\mathrm{120 \div 6 = 20}\) groups with only juniors

Using combination notation: \(\mathrm{C(6,3) = 20}\)

5. Apply complementary counting

Now we can find our answer:

Groups with at least one senior = Total groups - Groups with only juniors
Groups with at least one senior = \(\mathrm{120 - 20 = 100}\)

Process Skill: APPLY CONSTRAINTS - Using the complement to efficiently handle the "at least one" constraint.

Final Answer

The number of different groups of 3 partners that can be formed with at least one senior partner is 100.

This matches answer choice B. 100.

Verification: Our complementary counting approach avoided the complexity of calculating multiple cases while ensuring we counted every valid group exactly once.

Common Faltering Points

Errors while devising the approach

1. Misinterpreting "at least one senior partner"

Students often confuse "at least one senior" with "exactly one senior." They might think they only need to calculate groups with exactly 1 senior partner and 2 junior partners, missing the cases where there are 2 seniors + 1 junior, or 3 seniors + 0 juniors. This leads to an incomplete count and wrong answer.

2. Not recognizing the complementary counting opportunity

Many students attempt to solve this by calculating each case separately (1 senior + 2 juniors, 2 seniors + 1 junior, 3 seniors + 0 juniors) and then adding them up. While this approach works, it's more time-consuming and error-prone. Students miss the elegant complementary approach of subtracting "groups with no seniors" from "total groups."

3. Forgetting that order doesn't matter in group selection

Students sometimes approach this as a permutation problem instead of a combination problem. They might think that selecting "Senior A, Junior B, Junior C" is different from "Junior B, Senior A, Junior C" when forming groups, leading them to use permutation formulas instead of combination formulas.

Errors while executing the approach

1. Arithmetic errors in combination calculations

When calculating \(\mathrm{C(10,3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}}\), students often make basic arithmetic mistakes. They might incorrectly calculate \(\mathrm{10 \times 9 \times 8 = 720}\) as something else, or make errors when dividing by \(\mathrm{3! = 6}\), leading to wrong intermediate results.

2. Incorrect application of combination formula

Some students struggle with the combination formula itself. They might forget to divide by the factorial of the number being chosen (\(\mathrm{3!}\) in this case), or they might incorrectly apply the formula as \(\mathrm{C(10,3) = \frac{10!}{3! \times 7!}}\) by trying to use the full factorial formula but making computational errors.

Errors while selecting the answer

1. Calculating the complement incorrectly

After finding that total groups = 120 and groups with only juniors = 20, students might accidentally add these numbers instead of subtracting, getting \(\mathrm{120 + 20 = 140}\) instead of \(\mathrm{120 - 20 = 100}\). This type of final-step error is common when students lose track of what the complement represents.