The manager of a certain advertising agency must divide the group of 6 employees into a pair of teams with...

1. Translate the problem requirements: We need to understand that dividing 6 employees into two teams of 3 each means that once we choose 3 people for the lead team, the remaining 3 automatically form the backup team. The question asks specifically for the number of different lead teams possible.
2. Recognize the selection structure: This is a combination problem where we're selecting 3 people out of 6 for the lead team, and the order within the team doesn't matter.
3. Apply the combination formula: Calculate \(\mathrm{C(6,3)}\) to find the number of ways to choose 3 people from 6 for the lead team.
4. Verify the logic: Confirm that our answer makes sense by checking that each selection of 3 people for the lead team creates exactly one unique division of the group.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked. We have 6 employees total, and we need to split them into two teams of 3 people each. One team will be called the "lead team" and the other will be called the "backup team."

Here's the key insight: once we choose 3 people for the lead team, the remaining 3 people automatically become the backup team. There's no separate choice to make for the backup team.

For example, if our 6 employees are Alice, Bob, Charlie, Diana, Eve, and Frank:

• If we choose Alice, Bob, and Charlie for the lead team, then Diana, Eve, and Frank must be the backup team
• If we choose Alice, Diana, and Eve for the lead team, then Bob, Charlie, and Frank must be the backup team

So the question is really asking: "In how many different ways can we choose 3 people out of 6 for the lead team?"

Process Skill: TRANSLATE - Converting the problem language about "dividing into teams" into the mathematical understanding of "selecting members for one team"

2. Recognize the selection structure

Now let's think about what type of selection this is. When we're choosing 3 people for the lead team:

• The order doesn't matter (Alice-Bob-Charlie is the same lead team as Charlie-Alice-Bob)
• We can't choose the same person twice
• We're selecting a subset of people from a larger group

This is exactly what we call a "combination" - selecting items from a group where order doesn't matter.

Let's verify with a simple example. If we had only 4 people (Alice, Bob, Charlie, Diana) and wanted to choose 2 for the lead team, our options would be:

• Alice and Bob (leaving Charlie and Diana as backup)
• Alice and Charlie (leaving Bob and Diana as backup)
• Alice and Diana (leaving Bob and Charlie as backup)
• Bob and Charlie (leaving Alice and Diana as backup)
• Bob and Diana (leaving Alice and Charlie as backup)
• Charlie and Diana (leaving Alice and Bob as backup)

That's 6 different lead teams possible from 4 people.

3. Apply the combination formula

Now we can calculate this systematically. When we want to choose 3 people from 6 people where order doesn't matter, we use the combination formula.

In plain English, this means: "How many ways can we arrange 6 people in groups, then pick the first 3, accounting for the fact that the order of those 3 doesn't matter?"

The calculation works like this:

• First, count all possible ways to arrange 3 people chosen from 6: \(\mathrm{6 \times 5 \times 4 = 120}\)
• But this counts each team multiple times because Alice-Bob-Charlie is the same team as Bob-Alice-Charlie, Charlie-Alice-Bob, etc.
• Each group of 3 people can be arranged in \(\mathrm{3 \times 2 \times 1 = 6}\) different orders
• So we divide by 6 to avoid overcounting: \(\mathrm{120 \div 6 = 20}\)

Using the combination notation: \(\mathrm{C(6,3) = \frac{6!}{3! \times 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = \frac{120}{6} = 20}\)

4. Verify the logic

Let's double-check our reasoning. We found that there are 20 different ways to choose 3 people for the lead team from 6 total people.

This makes sense because:

• Each time we choose 3 people for the lead team, we automatically determine who's on the backup team (the remaining 3)
• Each selection creates exactly one unique division of the group
• We're not double-counting because we're only counting lead teams, not both lead and backup teams

If we were to count both lead teams AND backup teams separately, we'd get 40 total teams, but since the question specifically asks for lead teams only, our answer is 20.

Final Answer

The number of different lead teams possible is 20.

Looking at our answer choices:

1. 120
2. 60
3. 40
4. 20
5. 10

Our answer matches choice D. 20.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the constraint that teams must be distinct roles

Students may think they need to count all possible ways to divide 6 people into two groups of 3, not recognizing that the question specifically asks only for lead teams. This leads them to calculate \(\mathrm{C(6,3) \times C(3,3) = 20 \times 1 = 20}\), but then incorrectly think they need to account for swapping lead and backup roles, leading to wrong approaches.

2. Confusing combinations with permutations

Students may think the order within the team matters (like assigning specific roles within the lead team), leading them to use permutation formulas \(\mathrm{P(6,3) = \frac{6!}{(6-3)!} = 120}\) instead of the correct combination formula \(\mathrm{C(6,3) = 20}\).

3. Double-counting by considering both team assignments

Students may incorrectly think they need to count both "ways to form the lead team" AND "ways to form the backup team" separately, not realizing that once you choose 3 people for the lead team, the backup team is automatically determined.

Errors while executing the approach

1. Arithmetic errors in combination calculation

When calculating \(\mathrm{C(6,3) = \frac{6!}{3! \times 3!}}\), students may make errors like: incorrectly computing \(\mathrm{6 \times 5 \times 4 = 120}\), or making mistakes in \(\mathrm{3! = 6}\), or errors in the final division \(\mathrm{120 \div 6 = 20}\).

2. Using wrong combination formula

Students may use \(\mathrm{C(6,2)}\) instead of \(\mathrm{C(6,3)}\), or confuse which number goes where in the combination formula, leading to incorrect calculations like \(\mathrm{C(3,6)}\) which is undefined.

Errors while selecting the answer

1. Selecting the total arrangements instead of combinations

Students who calculated \(\mathrm{6 \times 5 \times 4 = 120}\) (the numerator of the combination formula) might select answer choice A (120) without completing the division by 3! to account for order not mattering.

2. Selecting the total number of teams instead of just lead teams

Students who correctly calculated that there are 20 different ways to select 3 people, but then reasoned "this creates 20 lead teams and 20 backup teams, so 40 total teams" might incorrectly select answer choice C (40) instead of D (20), misunderstanding that the question asks specifically for lead teams only.