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# Intermediate Permutation method/question

Author Message
Manager
Joined: 21 Feb 2008
Posts: 87

Kudos [?]: 22 [1], given: 2

Schools: UCLA Anderson '12

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14 Oct 2008, 16:30
1
KUDOS
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I've learned a few ways of solving these from the Manhattan GMAT book but wanted someone's opinion on which way is the best or most appropriate for a certain question. Would appreciate any of your thoughts. Thanks!

Example is something like: L, M, N, O, P are sitting next to each other. O and P do not want to sit next to each other so how many total permutations are allowed.

1st way:
Multiply the possibilities for each person, like a modified factorial:

P has 5 possibilities
O has 3 choices 2/5 of the time and 2 choices 3/5 of the time = 12/5 possibilities
N has 3 possibilities
M has 2 possibilities
L has 1 possibility
------------------------------
total = 5 x 12/5 x 3 x 2 x 1 = 72 possibilities

2nd way:
Count number of permutations O and P will sit next to each other manually. Seats 1&2, 2&1, 2&3, 3&2, 3&4, 4&3, 4&5, 5&4 -> 8 total. The other 3 people have 3! permutations of seating arrangements still so that's 8 x 3! = 48 permutations not allowed.

5! total permutations = 120 - 48 not allowed = 72 allowed

3rd way
Count number of allowed permutations per set. I've just shifted one letter (P) down each spot, so for this one set, 3/5 of the permutations are allowed. 3/5 x 5! total permutations = 72 allowed

PLMNO - ok
LPMNO - ok
LMPNO - ok
LMNPO - not ok
LMNOP - not ok

Kudos [?]: 22 [1], given: 2

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Intern
Joined: 23 Sep 2008
Posts: 15

Kudos [?]: [0], given: 0

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14 Oct 2008, 23:48
I think the easy way to do it is as such:

COunt all possible ways of arranging people ie; 5!
Then consider P and O to be one person and find the ways of arranging 4 people instead of 5. Therefore 4!.

But the only thing is even though P and O is one person, they can flip positions. Hence it will be 4!X2

Hence answer is 5!-4!X2 = 120 - 48 = 72

Kudos [?]: [0], given: 0

Manager
Joined: 01 Jan 2008
Posts: 222

Kudos [?]: 62 [0], given: 2

Schools: Booth, Stern, Haas

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15 Oct 2008, 00:39
duuuma wrote:
I've learned a few ways of solving these from the Manhattan GMAT book but wanted someone's opinion on which way is the best or most appropriate for a certain question. Would appreciate any of your thoughts. Thanks!

Example is something like: L, M, N, O, P are sitting next to each other. O and P do not want to sit next to each other so how many total permutations are allowed.

1st way:
Multiply the possibilities for each person, like a modified factorial:

P has 5 possibilities
O has 3 choices 2/5 of the time and 2 choices 3/5 of the time = 12/5 possibilities
N has 3 possibilities

can someone elaborate on that part?

Kudos [?]: 62 [0], given: 2

Manager
Joined: 21 Feb 2008
Posts: 87

Kudos [?]: 22 [0], given: 2

Schools: UCLA Anderson '12

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17 Oct 2008, 14:13
Sure -
If P sits in the outside seat (2 of 5 seats) -- (P x x x x) or (x x x x P) -- O can sit in 3 available seats in order to avoid sitting next to P.

If P sits in any of the middle seats (3 of 5 seats) -- (x P x x x) (x x P x x) (x x x P x) -- O can sit in 2 available seats in order to avoid sitting next to P.

kazakhb wrote:
duuuma wrote:
I've learned a few ways of solving these from the Manhattan GMAT book but wanted someone's opinion on which way is the best or most appropriate for a certain question. Would appreciate any of your thoughts. Thanks!

Example is something like: L, M, N, O, P are sitting next to each other. O and P do not want to sit next to each other so how many total permutations are allowed.

1st way:
Multiply the possibilities for each person, like a modified factorial:

P has 5 possibilities
O has 3 choices 2/5 of the time and 2 choices 3/5 of the time = 12/5 possibilities
N has 3 possibilities

can someone elaborate on that part?

Kudos [?]: 22 [0], given: 2

Re: Intermediate Permutation method/question   [#permalink] 17 Oct 2008, 14:13
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