Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many...

1. Translate the problem requirements: We need to divide 6 students into 3 equal groups (2 students each) and then assign each group to a different topic. We're counting total possible arrangements.
2. Form groups of students: Calculate how many ways we can divide 6 students into 3 groups of 2 students each, accounting for the fact that groups themselves are indistinguishable at this stage.
3. Assign topics to groups: Determine how many ways we can assign 3 different topics to the 3 groups we've formed.
4. Apply multiplication principle: Combine the results from group formation and topic assignment to get the total arrangements.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what this problem is asking us to do step by step.

We have 6 students total. We need to divide them into 3 equal groups - that means each group will have 2 students (since \(6 ÷ 3 = 2\)).

Then, we assign each of these 3 groups to a different topic. Since the topics are different, it matters which group gets which topic.

We want to count all the different ways this entire process can happen.

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

2. Form groups of students

Now let's figure out how many ways we can divide 6 students into 3 groups of 2 students each.

Let's call our students A, B, C, D, E, and F to make this concrete.

For the first group, we need to choose 2 students out of 6. That's \(6 × 5 ÷ 2 = 15\) ways (we divide by 2 because the order within a group doesn't matter - picking A then B is the same as picking B then A).

For the second group, we choose 2 students from the remaining 4. That's \(4 × 3 ÷ 2 = 6\) ways.

For the third group, the last 2 students automatically form the group. That's 1 way.

So far we have \(15 × 6 × 1 = 90\) ways.

But wait! We've overcounted because the order in which we form the groups doesn't matter at this stage. For example, forming groups (A,B), (C,D), (E,F) is the same as forming groups (C,D), (A,B), (E,F) - it's just a different order of the same three groups.

Since we can arrange 3 groups in \(3! = 6\) ways, we've counted each unique grouping 6 times.

So the actual number of ways to form the groups is \(90 ÷ 6 = 15\).

3. Assign topics to groups

Now we have our 3 groups of students formed. We need to assign 3 different topics to these 3 groups.

Since the topics are different (let's call them Topic X, Topic Y, and Topic Z), it matters which group gets which topic.

The first group can be assigned any of the 3 topics.
The second group can be assigned any of the remaining 2 topics.
The third group gets the last remaining topic.

So there are \(3 × 2 × 1 = 6\) ways to assign the topics to the groups.

4. Apply multiplication principle

To find the total number of different arrangements, we multiply the number of ways to form groups by the number of ways to assign topics.

Total arrangements = (Ways to form groups) × (Ways to assign topics)
Total arrangements = \(15 × 6 = 90\)

Final Answer

The total number of different arrangements possible is 90.

This matches answer choice (C) 90.

Common Faltering Points

Errors while devising the approach

Students might think "equally divided into 3 groups" means each group can have any number of students, rather than recognizing that 6 students divided equally into 3 groups means exactly 2 students per group. This leads to attempting more complex calculations with variable group sizes.

Students may miss that the topics are "different" and treat them as indistinguishable, thinking only about forming groups without considering that assigning Group 1 to Topic A is different from assigning Group 1 to Topic B.

Students might try to solve this as a single permutation problem, attempting to directly assign students to topic-group combinations, rather than breaking it down into: (1) form groups, then (2) assign topics to groups.

Errors while executing the approach

The most critical error is calculating \(15 × 6 × 1 = 90\) ways to form groups but forgetting to divide by \(3! = 6\) to account for the fact that the order of forming groups doesn't matter. This leads to getting \(90 × 6 = 540\) as the final answer.

Students may incorrectly calculate \(\mathrm{C}(6,2) = \frac{6!}{2! × 4!}\) as something other than 15, or make errors when computing \(\mathrm{C}(4,2) = 6\), leading to wrong intermediate results.

Some students might divide by 3! at the wrong step, such as dividing the final answer by 6, or might divide each individual combination by 6 rather than dividing the total number of group formations.

Errors while selecting the answer

If students forget to divide by 3! when forming groups, they get 90 ways to form groups, then multiply by 6 ways to assign topics, arriving at 540 (choice E) instead of the correct 90.

Students might correctly calculate 15 ways to form groups but forget about assigning topics entirely, potentially selecting 30 (choice A) if they make some other minor error, missing that they need to multiply by 6 for topic assignment.