In how many ways can $6$ boys and $6$ girls be seated at $3$ groups of four desks if there are two adjacent boys and two adjacent girls in each group?

Question

So the problem is,

There is a class of 12 students, 6 boys and 6 girls. There are 3 clusters of desks in a room. Each cluster consists of 2 boys sat next to each other and 2 girls sat next to each other. What is the total number of ways the students can be arranged.

This is what I have tried so far but not sure if I am right by using combinations: $$\dfrac{\left({6 \choose 2}\times{6 \choose2} \right)^3}{12!}$$ $$=\dfrac{225^3}{479001600}$$

I have drawn a picture of the problem to clarify.

Answer 1

Strategy:

1. Choose which two of the six boys and which two of the six girls sit at the cluster of four desks closest to Mr. Zerk.
2. Choose on which side of that cluster of desks the boys sit.
3. Arrange the two selected boys on that side.
4. Arrange the two selected girls on the other side.
5. Choose which two of the four remaining boys and which two of the four remaining girls sit at the cluster of four desks closest to the sliding door.
6. Choose on which side of that cluster of desks the boys sit.
7. Arrange the two selected boys on that side.
8. Arrange the two selected girls on the other side.
9. Choose on which side of the remaining cluster of four desks the remaining two boys will sit.
10. Arrange those boys on that side.
11. Arrange the two remaining girls on the other side.
12. Apply the Multiplication Principle.

$$\binom{6}{2}\binom{6}{2}\binom{2}{1}2!2!\binom{4}{2}\binom{4}{2}\binom{2}{1}2!2!\binom{2}{1}2!2!$$