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$6$ people spread in $3$ distinguishable rooms, every room needs one person who opens the door.

There are ${6 \choose 3}\cdot 3 \cdot 2$ options to choose the three door opener persons and let them open one certain room, so that every room is opened by one person.

Further, there are then $3$ options to have all left three guys in one of the rooms, $3\cdot 2$ options to have exactly one other guy in every room, and ${3 choose 2}\cdot 3 \cdot 2$ options to have one guy in one room, two in the other and none in the last room.

So in total there are ${6 \choose 3}\cdot 3 \cdot 2 \cdot 3 \cdot 3 \cdot 2 {3 \choose 2} \cdot 3 \cdot 2$ options to have all rooms opened by exactly one person and spread the others in these rooms.

If only there are 3 people who can open the doors (only they have keys for all the rooms). There are only $3 \cdot 2 \cdot 3 \cdot 3 \cdot 2 {3 \choose 2} \cdot 3 \cdot 2$ options left, right?

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Could you explicitly state the question? Are you looking for the number of different ways we can put $6$ distinguishable people into $3$ distinguishable rooms? – Omnomnomnom Jul 22 '13 at 17:59
Yes, but under the constraint, that the rooms are distinguishable and one per room is "the door opener" – combizombi Jul 22 '13 at 18:05

Here's an easy way to look at the problem:

First, we make sure that we have a door opener in each room. There are $6$ choices for the first room, $5$ for the second, and $4$ for the third. Thus, we may select door openers in $6\times5\times4=120$ ways.

Now, each of the three remaining people can be put in any of the three rooms. So, we have $3\times3\times3=27$ ways of distributing the remaining people after the door openers are chosen.

Thus, the total number of possibilities is $120\times27=3240$

As far as I can tell, you've followed the same steps, but I think this is a bit less redundant. For example, instead of writing that there are $\binom{6}{3}\times3\times2$ ways of selecting door openers, you could have noticed that $$ \begin{align} \binom{6}{3}\times3\times2 &= \frac{6!}{3!3!}\times3!\\ &=\frac{6!}{3!}\\ &=\frac{6\times5\times4\times3\times2}{3\times2}\\ &=6\times5\times4 \end{align} $$

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Thanks a lot! Seems much simpler ;-) And what about the slightly altered problem, when only 3 people are available as door openers? Distributing the remaining people stays the same, but instead of 6*5*4 ways to select door openers there are only 3*2*1, right? – combizombi Jul 22 '13 at 22:43
Exactly. Total is then $6\times 27=162$ – Omnomnomnom Jul 23 '13 at 4:54

If you choose an order for the six people and put the first two into room $1$ the next two into room $2$ and the final pair into room $3$, and then you put the first named of each pair as the door-opener, every one of the $6!$ orders of people gives you a unique arrangement of people in accordance with the criteria in the problem. And from the arrangement of people, you can recover a unique order for the six people.

So I reckon you should be looking at $6!=720$ possibilities. The factor $5$ comes from your $\binom 63$ which seems to go missing somewhere.

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But then in every room there is an equal number of people? – combizombi Jul 22 '13 at 18:33
I thought the question had two in each room - I was going to edit this, but Omnomnomnom put almost exactly what I was going to write. – Mark Bennet Jul 22 '13 at 20:37

Answer would be different if interpreted like this. Instead of designating a person as a "door-man" if it were that any-person inside the room could open the door (and ofcourse there has to be someone inside to open the room) The solution would be as follows.

3^6 - (possibilities one or two rooms could be empty) Exactly 2 rooms could be empty = Just 3 ways. Exactly 1 room could be empty = 3C1 * (2^6 - 2) = 186 ways.

Therfore the asnwer is 729 - 3 - 186 = 540 ways.

Please confirm this.

Warm Regards, Shanky

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