1. ## Permutation/combination

there are 7 people
1 room for 3,
1 room for 2 and
1 room for 1 (1 will have to sleep in the car)
how many different arrangements are there?

2. Let's make the question somewhat simpler to begin with. If the situation was like this: 1 room for 3 people, the rest will sleep outside. Then how many different arrangements are there?

If you can answer that, then think about the additional number of arrangements we get by dividing the remaining 4 amongst two other rooms and a car (adding 1 room or car at a time).

3. Originally Posted by Mobius
Let's make the question somewhat simpler to begin with. If the situation was like this: 1 room for 3 people, the rest will sleep outside. Then how many different arrangements are there?

If you can answer that, then think about the additional number of arrangements we get by dividing the remaining 4 amongst two other rooms and a car (adding 1 room or car at a time).
My answer was 7! but I think its wrong x.x

4. is it 7C3 x 4C2 x 2C1 x 1C1x 4! ?

5. Almost! Just remove the 4! factor at the end.

6. Hello, ozacho!

There are 7 people,
1 room for 3 people, 1 room for 2 people and 1 room for 1 person.
(One will have to sleep in the car)

How many different arrangements are there?

This is an ordered partition.

There are: . $\displaystyle {7\choose3,2,1,1} \:=\:\frac{7!}{3!\,2!\,1!\,1!} \:=\:420$ arrangements.

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We can baby-step through the reasoning . . .

There are 7 people.
Choose 3 of them to go in the three-person room.
. . There are: . $_7C_3 \:=\:35$ choices.

Of the remaining 4 people, choose 2 to go in the two-person room.
. . There are: . $_4C_2 \:=\:6$ choices.

Of the remaining 2 people choose 1 to go in the one-person room.
. . There are: . $_2C_1 \:=\:2$ choices.

And the 1 remaining person will sleep in the car.
. . There is: . $1$ choice.

Therefore, there are: . $35\cdot 6\cdot 2\cdot 1 \:=\:420$ choices (arrangements).