Problem Solving With Combinations -- A Hard One!

Question

A cabin has 2 rooms with 3 single beds each, 1 room with 2 single beds, and 1 room with 4 beds. 6 girls and 6 boys are assigned to rooms with people of the same gender. In how many ways can the rooms be assigned.

I've tried everything guys, but I can't get the answer, which is 600. My problem is making sure the same gender stays in one room. Just don't know how to do it.

A tip for you trying to answer questions, nCr, nPr is the format I'm looking for, not functions (ie n(r)).

Answer 1

If the girls get 4,2 and the boys get 3,3, then there are $6C2*6C3=15*20=300$ ways (pick 2 girls for the 2-room and pick 3 boys for the first 3-room). Similarly, there are $300$ ways if the genders are reversed, for a total of $600$.

Answer 2

Let's break it up.

Room $1$: $3$ beds

Room $2$: $3$ beds

Room $3$: $2$ beds

Room $4$: $4$ beds

We have a total of $6$ girls and $6$ boys

There's really only two ways we can break this up:

Boys can go in rooms $1/2$, and girls go in rooms $3/4$

Or, boys go in rooms $3/4$, and girls go in rooms $1/2$

Since there are an equal number of boys and girls, we can solve it for one gender, and then just times by $2$.

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If the boys get rooms $1/2$, that means they must be split into $\text{1 1 1 2 2 2}$, where those are the rooms. We can do this a total of $6C3$ ways. So $3$ boys will go in room $1$, and the remaining will go in room $2$. Thus $6C3=20$

If the boys get rooms $3/4$, that means they must be split into $\text{3 3 4 4 4 4}$, where those are the rooms. We can do this in a total of $6C2=6C4$ ways. So $2$ boys will go in room $3$, and the remaining will go in room $4$. Thus $6C2=6C4=15$

Since either one of these scenarios is possible and independent, we can multiple giving us $20*15=300$, and of course, multiply by $2$ for the girls, thus $600$

Answer 3

To comply with the same gender rule, either boys occuppy the 2 rooms with 4 and 2 beds respectively (girls take the rest) or the other way around. hence:

Total ways=$2\times$ A(ways to arrange 6 people in 2 rooms of 4 and 2 beds respectively) $\times$ B(ways to arrange 6 persons in two room each with 3 beds)

$A=\binom{4}{6}=\binom{2}{6}=6\times 5\div 2=15$ (Pick 4 for one room and the rest take the other)

$B=\binom{3}{6}=6\times 5\times 4/(3\times 2)=20$ (Pick 3 for one room and the rest take the other)

And total ways = $2\times 15\times 20=600$