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

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Correct

180

540

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Explanation

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Official Explanation

Official

First we need to select the groups: Since we have 6 students, the first group can be formed in

6C2 = \frac{6 \times 5}{2!} = 15

ways.

Since there are now 4 students left, the 2nd group can be selected in

4C2 = \frac{4 \times 3}{2!} = 6

ways.

Since there are 2 students left, the final group can be selected in

2C2 = 1

way.

Thus, the total number of ways to select the 3 groups is

15 \times 6 \times 1 = 90

if the order of selecting these groups matters.…

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What this question tests

Whether you can count arrangements and selections correctly — permutations vs. combinations.

Overcounting: treating a selection as ordered when order does not matter, or forgetting to divide by duplicates.