Math Help - Combinations

1. Combinations

The questions:

1) There are 8 free rooms in a hostel. How many ways can 4 students be divided into these rooms?

2) A student must choose 5 question from a 12 question exam paper. If he chooses at least 3 from the first 6, calculate the number of ways he can choose.

1) Since order does not matter, we calculate as such: $\left(^{8}\mathcal{P}_5\right)$ which will give us 5040. Is this right or am I doing something wrong?

2)Choosing at least 3 from the first 6 it will give 12X11X10 = 1320. From here, I am stuck assuming I was right upto here in the first place.

2. Originally Posted by Keep
The questions:
1) There are 8 free rooms in a hostel. How many ways can 4 students be divided into these rooms?
2) A student must choose 5 question from a 12 question exam paper. If he chooses at least 3 from the first 6, calculate the number of ways he can choose.
1) Since order does not matter, we calculate as such: $\left(^{8}\mathcal{P}_5\right)$ which will give us 5040. Is this right or am I doing something wrong?
2)Choosing at least 3 from the first 6 it will give 12X11X10 = 1320. From here, I am stuck assuming I was right upto here in the first place.
In #1 order does make a difference because presumably the rooms are numbered.
So it is a permutation and that the notation you used but the wrong answer.
$\left(^{8}\mathcal{P}_4\right)=(8)(7)(6)(5)=1680$

For the second question it is a combination: $\left(^{6}\mathcal{C}_3\right)\cdot\left(^{9}\math cal{C}_2\right)$.

3. Originally Posted by Plato
In #1 order does make a difference because presumably the rooms are numbered.
So it is a permutation and that the notation you used but the wrong answer.
$\left(^{8}\mathcal{P}_4\right)=(8)(7)(6)(5)=1680$

For the second question it is a combination: $\left(^{6}\mathcal{C}_3\right)\cdot\left(^{9}\math cal{C}_2\right)$.
Thank you Plato. In number one, the answer will be 1680 if we presume that the rooms are numbered and hence that order matters. But what if we assume that the rooms are not numbered and there is no order. Then it will be $\left(^{8}\mathcal{C}_4\right)$ which is equal to 56, right?

PS:Sorry I meant to write C, not P before.

4. Originally Posted by Keep
But what if we assume that the rooms are not numbered and there is no order. Then it will be $\color{red}\left(^{8}\mathcal{C}_4\right)$ which is equal to 5040, right?
No indeed. $\left(^{8}\mathcal{C}_4\right)=\frac{8!}{(4!)(4!)} =70$

5. Originally Posted by Plato
No indeed. $\left(^{8}\mathcal{C}_4\right)=\frac{8!}{(4!)(4!)} =70$

Sorry it is all my mistakes. Nevermind. I get it now. I thought I wrote 5 rooms and then did not calculate properly.