A variation of a combination and a permutation, I think? The scenario is that 6 people have the option of choosing 8 doors and we want to know each door a person goes through. I have four/five questions based on this.
1) How many different ways can 6 people go into the doors? How many if each go through a different door?
For the first part I was thinking $6^8$ because each person has 8 options and there is 6 people. For the second one, I was thinking $8!/6!$.
2)How many ways can they pick a door where at least 2 pick the 3rd door. 
For this one I think its logical just to assume they are going for the first door in terms of a mapping, so just imagine the question is at least 2 pick the 1st door. If at least 2 pick the first door, then have 4 people left with 7 doors. Granted they could all go to the first door as well. This is where I get confused thought.
3) How many ways can they pick a door such that 4  pick one door and 2 pick another door?
I thought that well if 4 pick the first door and 2 pick the second, then if we keep moving the 2 people down the line we get a total of 7 ways that time. Then we can move it again and again etc. So there are 8 ways to get this, hence we have $7*8=56$ ways.
4) How many ways can 2 people pick three different doors, so 2 people pick the same door 3 times. (If this doesnt make sense I will do my best to clarify it more).
This one, I am a little lost on how to begin. 
 A: 1)
subpart i) Each person has 8 doors to choose from, and there are 6 people. so the answer is $8\cdot 8 \cdot 8 \cdot 8 \cdot 8 \cdot 8 = 8^6$
subpart ii) Each person has to choose a different door, so the first person has 8 options, the second 7, the third 6... The answer is now $8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 = 20160$
2)
You're right, we can simply fix the first two people to go through the third door, but then the rest must pass through other doors. It's easier to look at all the possibilities where less than two people choose the third door and subtract those from all possible combinations. In $7^6$ combinations no one uses the third door, and in $8 \cdot 1 \cdot 7^5$ only one person passes through the door. So the final answer is $8^6 - 7^6 - 8 \cdot 7^5$
3) The question doesn't specify that four people must choose the first door, but any one door. So let's start by choosing the four people out of six total that will go through one door (there are 8 choices total, so we multiply by 8), and then the remaining two can go through some other door (7 other doors left, so we multiply by 7). The final answer should be:
${6 \choose 4} \cdot 8 \cdot 7 = 840$
I had an incorrect answer before, I forgot to account for different doors for the second person.
I don't really understand part 4. It seems there are two people total and three doors. But how many times are they picking a door?
