Six students are equally divided into 3 groups, then, the three groups

First we need to select the groups:

Since we have 6 students, the first group can be formed in 6C2 = (6 x 5)/2! = 15 ways. Since there are now 4 students left, the 2nd group can be selected in 4C2 = (4 x 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 x 6 x 1 = 90 if the order of selecting these groups matters. However, the order of the selection doesn’t matter, so we have to divide by 3! = 6. Thus, the total number of ways to select the groups when order doesn’t matter is 90/6 = 15.

Now we need to determine the number of ways to assign those 3 groups to 3 different topics, which can be done in 3! = 6 ways.

Thus, the total number of ways to form the groups and assign them to a task is:

15 x 6 = 90

Answer: C