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

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A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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A certain company assign 5 rooms to each of its 5 employees in marketing department. Next month, Company plans to change these employees rooms in such a way that no employee get the same room. In how many ways company can assign rooms to these employees.

a) 5
b) 8
c) 44
d) 60
e) 120

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Originally posted by sananoor on 27 Jul 2017, 04:25.
Last edited by sananoor on 27 Jul 2017, 07:36, edited 2 times in total.
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 06:04
Let A, B, C, D, E be the employees, and their respective assigned rooms initially are R1, R2, R3, R4, R5.

In order for the employees to not get the same room every employee has 4 options (5 choices - 1, his old room).

Rooms can be assigned in the following manner:
A: R2, R3, R4, R5.
B: R1, R3, R4, R5.
C: R1, R2, R4, R5.
D: R1, R2, R3, R5.
E: R1, R2, R3, R4.

So, rooms can be assigned in 4+4+4+4 = 16 ways. Ans - C.
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 06:36
TimeTraveller wrote:
Let A, B, C, D, E be the employees, and their respective assigned rooms initially are R1, R2, R3, R4, R5.

In order for the employees to not get the same room every employee has 4 options (5 choices - 1, his old room).

Rooms can be assigned in the following manner:
A: R2, R3, R4, R5.
B: R1, R3, R4, R5.
C: R1, R2, R4, R5.
D: R1, R2, R3, R5.
E: R1, R2, R3, R4.

So, rooms can be assigned in 4+4+4+4 = 16 ways. Ans - C.


Your answer matches with the but I don't understand your logic. How can we just add 4+4+4+4 when giving a room to one person is directly related to other. For e.g. if I assign R2 to A, we can't assign R2 to C, D, E . So, I don't think u can just add 4+4+4+4..


Also 4+4+4+4+4 for each A,B,C,D,E is 20 and not just 16..
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 06:43
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sananoor wrote:
A certain company assign 5 rooms to each of its 5 employees in marketing department. Next month, Company plans to change these employees rooms in such a way that no employee get the same room. In how many ways company can assign rooms to these employees.

a) 5
b) 8
c) 16
d) 60
e) 120


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
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A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post Updated on: 27 Jul 2017, 07:41
I am not getting this question
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Originally posted by sananoor on 27 Jul 2017, 06:45.
Last edited by sananoor on 27 Jul 2017, 07:41, edited 2 times in total.
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 06:47
chetan2u wrote:
sananoor wrote:
A certain company assign 5 rooms to each of its 5 employees in marketing department. Next month, Company plans to change these employees rooms in such a way that no employee get the same room. In how many ways company can assign rooms to these employees.

a) 5
b) 8
c) 16
d) 60
e) 120


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


Chetan can you see my explanation and let me know if i am wrong or not?
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 06:48
sananoor wrote:
Let me explain this question
suppose the 5 employees are A, B, C, D and E and five rooms assigned to these 5 employees are 1,2,3,4, and 5. Next months company want to assign them different rooms.
Now if A is assigned to room 1, then next month A has 4 options to select the room 2,3,4, and 5. (4 options)
B will also have 4 options. B can take either room 1 (assigned to A this month) or rooms C, D and E (4 options)
Now A and B are assigned with 2 rooms, three rooms are left.
C cant take the same room again and A and B are already assigned with 2 rooms. So out of 5, C has only 2 options left (2 options)
D will have only one option as D cant take the same room and 3 rooms are already taken by A, B and C...therefore, out of 5 D is left with one room (1 option)
E in the last will only be left with 1 last room. (1 option)
multiply all options
4*4*2*1*1 = 16 C is answer


4*4*2=32..
also you are taking it in a very simple way..
missing many ways
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 06:53
chetan2u wrote:
sananoor wrote:
Let me explain this question
suppose the 5 employees are A, B, C, D and E and five rooms assigned to these 5 employees are 1,2,3,4, and 5. Next months company want to assign them different rooms.
Now if A is assigned to room 1, then next month A has 4 options to select the room 2,3,4, and 5. (4 options)
B will also have 4 options. B can take either room 1 (assigned to A this month) or rooms C, D and E (4 options)
Now A and B are assigned with 2 rooms, three rooms are left.
C cant take the same room again and A and B are already assigned with 2 rooms. So out of 5, C has only 2 options left (2 options)
D will have only one option as D cant take the same room and 3 rooms are already taken by A, B and C...therefore, out of 5 D is left with one room (1 option)
E in the last will only be left with 1 last room. (1 option)
multiply all options
4*4*2*1*1 = 16 C is answer


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?
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 06:59
shashankism wrote:
TimeTraveller wrote:
Let A, B, C, D, E be the employees, and their respective assigned rooms initially are R1, R2, R3, R4, R5.

In order for the employees to not get the same room every employee has 4 options (5 choices - 1, his old room).

Rooms can be assigned in the following manner:
A: R2, R3, R4, R5.
B: R1, R3, R4, R5.
C: R1, R2, R4, R5.
D: R1, R2, R3, R5.
E: R1, R2, R3, R4.

So, rooms can be assigned in 4+4+4+4 = 16 ways. Ans - C.


Your answer matches with the but I don't understand your logic. How can we just add 4+4+4+4 when giving a room to one person is directly related to other. For e.g. if I assign R2 to A, we can't assign R2 to C, D, E . So, I don't think u can just add 4+4+4+4..


Also 4+4+4+4+4 for each A,B,C,D,E is 20 and not just 16..


Hi, yes it's a derangement problem, I didn't read it correctly I guess.
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 07:08
sananoor wrote:
chetan2u wrote:
sananoor wrote:
Let me explain this question
suppose the 5 employees are A, B, C, D and E and five rooms assigned to these 5 employees are 1,2,3,4, and 5. Next months company want to assign them different rooms.
Now if A is assigned to room 1, then next month A has 4 options to select the room 2,3,4, and 5. (4 options)
B will also have 4 options. B can take either room 1 (assigned to A this month) or rooms C, D and E (4 options)
Now A and B are assigned with 2 rooms, three rooms are left.
C cant take the same room again and A and B are already assigned with 2 rooms. So out of 5, C has only 2 options left (2 options)
D will have only one option as D cant take the same room and 3 rooms are already taken by A, B and C...therefore, out of 5 D is left with one room (1 option)
E in the last will only be left with 1 last room. (1 option)
multiply all options
4*4*2*1*1 = 16 C is answer


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?


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.
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 07:16
sananoor wrote:
chetan2u wrote:
sananoor wrote:
Let me explain this question
suppose the 5 employees are A, B, C, D and E and five rooms assigned to these 5 employees are 1,2,3,4, and 5. Next months company want to assign them different rooms.
Now if A is assigned to room 1, then next month A has 4 options to select the room 2,3,4, and 5. (4 options)
B will also have 4 options. B can take either room 1 (assigned to A this month) or rooms C, D and E (4 options)
Now A and B are assigned with 2 rooms, three rooms are left.
C cant take the same room again and A and B are already assigned with 2 rooms. So out of 5, C has only 2 options left (2 options)
D will have only one option as D cant take the same room and 3 rooms are already taken by A, B and C...therefore, out of 5 D is left with one room (1 option)
E in the last will only be left with 1 last room. (1 option)
multiply all options
4*4*2*1*1 = 16 C is answer


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...
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
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 07:33
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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.
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Re: A certain company assign 5 rooms to each of its 5 employees  [#permalink]

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New post 27 Jul 2017, 07:40
Answer edited and thanks Chetan and Ian
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Re: A certain company assign 5 rooms to each of its 5 employees   [#permalink] 27 Jul 2017, 07:40
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