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Intermediate Permutation method/question
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14 Oct 2008, 16:30
1
Kudos
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
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Re: Intermediate Permutation method/question
[#permalink]
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
Re: Intermediate Permutation method/question
[#permalink]
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
Re: Intermediate Permutation method/question
[#permalink]
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?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.