A certain company assign 5 rooms to each of its 5 employees

hi

there is a direct formula and it is called DERANGEMENT method, where none is its original position..
=n!-nC1*(n-1)!+nC2(n-2)!...
here n is 5..
5!-5C1*4!+5C2*3!-5C3*2!+5C4*1!-5C5= 120-120+60-20+5-1=44..

the Q does not have the correct answer..
Also it is not a GMAT type Q

I am not getting this question

Chetan can you see my explanation and let me know if i am wrong or not?

4*4*2=32..
also you are taking it in a very simple way..
missing many ways

thanks for letting me know where i am wrong
can u explain some easy way?

Hi, yes it's a derangement problem, I didn't read it correctly I guess.

It's a concept called probability derangement, or in this case simply derangement, another name is hat-check problems.

Basically, if there are n greeting cards and n envelopes with addresses written on them, the number of ways the greeting cards can be put inside the envelopes such that none of the greeting card was put inside the intended envelope = \(n! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + ... + (-1)^n \frac{1}{n!})\). In this case n = 5.

hi...
one is derangement ..
other would be to count the methods..
let me help with that..
lets freeze the rooms one by one..
1) A-b..
B can take a, so other three have only two ways C-d, D-e, E-c OR C-e,D-c, E-d.......2 ways
B takes any of three say c, now remaining three can take in three ways C-d, D-e, E-b OR C-e,D-b, E-d OR C-b,D-e, E-d.. 3 ways..
similarly three ways when B takes d and 3 ways when B takes e..

total = 2+3*3=11..
2)similarly for A-c or d or e..

total 11*4=44

Assuming we're assigning one employee per room, this is what is known, in advanced combinatorics, as a 'derangement' problem. The answer is 44, so is not among the answer choices. A complete solution is complicated (there are a few cases, none of them all that simple). This kind of problem is also well beyond the scope of the GMAT. What is the source?

I can at least show how you'd get started on the problem. We need to move the employee from room '1' to a new room, so we have 4 choices. Let's call that new room 'W'. Now the employee from room W has to move to a new room. Here's where you first have two cases - that employee could move to room 1, or could move to a room we haven't considered yet. If you just focus on the second case, where the employee from W goes to a new room (not room 1, but instead some other room 'X'), you then have 3 choices for that new room. And if you continue to assume that each employee goes to a room you haven't considered yet, you next have 2 choices, and finally 1 choice, for a total of 4*3*2*1 = 24 options. This is what we're really doing:

1 --> W (four choices)
W --> X (three choices)
X --> Y (two choices)
Y --> Z
Z --> 1

and since W, X, Y and Z need to be distinct room numbers from 2, 3, 4, and 5, there are 4! = 24 ways to choose their values.

But that's just one of three cases you need to analyze to solve this problem. You also have to count the case where employee 1 and employee X simply swap rooms, and then the cases where employee Y moves back to either room 1 or room X. If you add on those other two cases, you'll get 44 total possibilities. This is far too hard to be a real GMAT problem, so it is not worth worrying about.

Answer edited and thanks Chetan and Ian