Alice, Bob, Carol, Dave, Erin and Frank all live in the same dorm. Each has their own room. One night they all go out to the clubs and get wasted. Upon returning exactly two of them are able to find and sleep it off in their own rooms while each of the other four sleeps in a room not their own. (Nobody is sleeping together here) How many ways can such a scenario take place? (E.g., one way is Alice and Bob each sleep in their own room, whilst each of Carol, Dave, Erin and Frank sleep in a room other than their own.) Show both how to compute this and give a numeric answer.

Question

Alice, Bob, Carol, Dave, Erin and Frank all live in the same dorm. Each has their own room.  One night they all go out to the clubs and get wasted.  Upon returning exactly two of them are able to find and sleep it off in their own rooms while  each of the other four sleeps in a room not their own. (Nobody is sleeping together here)  How many ways can such a scenario take place?  (E.g., one way is Alice and Bob each sleep in their own room,  whilst each of Carol, Dave, Erin and Frank sleep in a room other than their own.)   Show both how to compute this and give a numeric answer.

anonymous · Answer

Yes indeed this is derangement problem.

anonymous · Answer

Would it be $$C(6,2) * !4$$ ?

And I don't know what the answer is.

anonymous · Answer

Looks good.

anonymous · Answer

The closed form is given by Rencontres numbers.

$$ D_{n,k} = {n \choose n-k} D_{n-k,0} = \frac{n!}{k!} \sum_{i=0}^{n-k} \frac{(-1)^i}{i!} $$

anonymous · Answer

Do yo know why it would be $$!4$$ and not $$!6$$

anonymous · Answer

The derivation is a bit trick, see here: http://math.stackexchange.com/questions/17320/

anonymous · Answer

Gotcha. Thanks!

anonymous · Answer

Glad to help :)

anonymous · Answer

mind=blown ;)