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#2 (permalink) Wed Jul 23, 2008 8:49 am permutation problem |
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. Sorry, I'm no good at math. Got a grammar question? . _________________ Native English teacher at Mister Micawber's |
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Mister Micawber Language Coach
Joined: 17 Jul 2005 Posts: 13015
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#3 (permalink) Wed Jul 30, 2008 16:02 pm permutation problem |
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1) She invited 4 friends, so there are 5 of them. To confirm this you are given the number 120
Permutation of 5*4*3*2*1= 120
So 4 friends and Mary always in the middle, you can diligently write out all the 24 different way..if M = Mary and her friends are ABCD, You can write
AB M CD BA M CD BA M DC AB M DC CD M AB
Keep going, you should get 24...time waster huh
OR you can simply think of it this way ...these approach is on 800score.com
5 places : _ _ M _ _
You have 4 people filling up those spaces beside Mary There first place you have a possible of 4 people who can fill that space, Second space you now have 3 people, on the other side of Mary, you have the 3 space, and 2 people choose from and finally one space to one person
therefore you have
4 3 M 2 1
Multiply all these together you get 4*3*2*1 = 24
Just sheer logic...If you can do the math approach just use logic in permutation problems as in writing every frigging thing out!! |
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Peekaboo New Member
Joined: 28 Jul 2008 Posts: 9
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#4 (permalink) Wed Jul 30, 2008 16:05 pm permutation problem |
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1) She invited 4 friends, so there are 5 of them. To confirm this you are given the number 120
Permutation of 5*4*3*2*1= 120
So 4 friends and Mary always in the middle, you can diligently write out all the 24 different way..if M = Mary and her friends are ABCD, You can write
AB M CD BA M CD BA M DC AB M DC CD M AB
Keep going, you should get 24...time waster huh
OR you can simply think of it this way ...these approach is on 800score.com
5 places : _ _ M _ _
You have 4 people filling up those spaces beside Mary There first place you have a possible of 4 people who can fill that space, Second space you now have 3 people, on the other side of Mary, you have the third space, and 2 people choose from and finally one and only one possible person to occupy that space
therefore you have
4 3 M 2 1
Multiply all these together you get 4*3*2*1 = 24
Just sheer logic...If you can do the math approach just use logic in permutation problems as in writing every frigging thing out!! |
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Peekaboo New Member
Joined: 28 Jul 2008 Posts: 9
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#5 (permalink) Wed Jul 30, 2008 16:19 pm permutation problem |
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sorry . i misunderstood the problem. i thought middle means any position in between 2 people. i dint think it meant exactly in the middle of the row. anyway thanks for clearing the confusion. |
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Abhi_86 I'm new here and I like it ;-)
Joined: 15 Feb 2007 Posts: 36
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#6 (permalink) Wed Jul 30, 2008 17:42 pm permutation problem |
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well I interpreted it that way in order to solve. it..But I just understood what you meant and its quite possible..One of the interesting thing about practicising this test and also LEARNING THE LANGUAGE..because regular folks like us will understood thing differently but ETS has it LANGUAGE..in the most part they try to be unequivocal as much as possible.
You can try using your idea and the permutations explanation above to see if you get the same answer, on the otherhand, someone might have a diferent explanation for it...you will probably come across it in Kaplan or any of the other ones or just google the questions and you might hit on someone who explained it elsewhere...there are folks on Yahoo Answers devoted to explaining GRE maths problems.. |
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Peekaboo New Member
Joined: 28 Jul 2008 Posts: 9
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#7 (permalink) Mon Aug 10, 2009 15:47 pm hi |
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| martha can sit in the middle...so fix the position....if martha has to be in the middle then there are remaining 4 positions which can be filled by four others in 4! =24 ways |
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Avinash Meda New Member
Joined: 10 Aug 2009 Posts: 1
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