In how many ways can a class of eight students be divided into four groups each having two students?

Question

Suppose there are eight students in a class. We have to divide them into four groups each having two students. The question is, in how many ways can it be done?

The answer is: $$\frac{8C2 \cdot 6C2 \cdot 4C2 \cdot 2C2}{4!}$$

I have a small confusion. Why do we need to divide by 4!?

Answer 1

A method to count this without needing to rely on dividing for symmetry:

Take your eight students and arrange them in some standard fashion. For the purpose of illustration, I will choose to do this by height, but other methods of ordering will work as well.

• Take the shortest person in the group. (there is no choice to be made here, it is predetermined).

• Pick a person not yet taken to be made the shortest person's group partner. (Seven choices available)

• From those six people remaining, take the shortest remaining person in the group. (again, there is no choice to be made here, it is predetermined based on previous choices)

• Pick a person still remaining to be made this person's group partner. (Five choices available)

• $\vdots$ Repeat the process, taking the current shortest person from those remaining and then picking a partner for them.

Applying multiplication principle, there are $1\cdot 7\cdot 1\cdot 5\cdot 1\cdot 3\cdot 1\cdot 1 = 7\cdot 5\cdot 3 = 7!! = 105$ (here $7!!$ is double factorial notation)

Note that this is equal to $\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}\frac{1}{4!}=\binom{8}{2,2,2,2}\frac{1}{4!}=\frac{8!}{(2!)^44!} = 105$

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Explaining the way the answer was written and how to view it that way,

Temporarily assign group names: group1, group2, group3, group4.

Arrange the eight people into these groups with 2 each in $\binom{8}{2,2,2,2}$ ways. (to see this, pick two to go to group1, then pick two remaining to go to group 2, etc... for a total of $\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$ ways)

Now, recognize that we have counted every scenario exactly $4!$ times each since each rearrangement of the group names yields the "same" outcome. Dividing by $4!$ corrects our count, making the total:

$\binom{8}{2,2,2,2}\frac{1}{4!}=105$