Suppose the set of students is
{A, B, C, D, E, F, G, H, I, J, K, L}
We will first do the problem as though
the three groups were ordered, then we
will divide by the number of different
orderings. To explain that:
Look at a sample grouping:
Group 1: {A, C, G, J}
Group 2: {E, I, K, L}
Group 3: {B, D, F, H}
But that is considered the same grouping as when we
re-label Group 1 as "Group 3", re-label Group 2 as "Group 1",
and re-label Group 3 as "Group 2", like this:
Group 1: {E, I, K, L}
Group 2: {B, D, F, H}
Group 3: {A, C, G, J}
The above two as well as any of the 3! or 6 ways to arrange
these three groups are considered the be the same grouping. So
once we have found the number of ways to divide the 12 into
one such set of groups, we will then have to divide by 3! or 6
1. Choose the students for Group 1 any of 12C4 ways.
That leaves 8 to choose Group 2 and Group 3 from.
2. For each of those 12C4 ways to choose Group 1, there are
8C4 ways to choose Group 2.
3. That leaves 4 students to choose
Group 3 from, and that means we choose all four of them,
which is the same as 4C4 (which is the same as 1 way).
So the answer to the first part is 12C4*8C4*4C4
But then we have to divide by 3! as explained earlier.
So the final answer is
, choice (a)
Edwin
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