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# M01 #24

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M01 #24 [#permalink]

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19 Nov 2009, 12:55
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Can someone give me an explanation on how to solve these probability in a circle type problems?

The formula is (N-1)! but why?

If 5 noble knights are to be seated at a round table, then how many different ways can they be seated?

120
96
60
35
24

[Reveal] Spoiler:
E
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Joined: 10 Jun 2009
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Re: M01 #24 [#permalink]

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20 Nov 2009, 01:48
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Now if we were to arrange them on a bench then we would use following formula n! i.e; 1st position can be taken by any 1 of the 5 knights, 2nd position can be taken by only 1 of the 4 knights because 1 knight has already taken the 1st position, simillarly 3rd position can be taken by any 1 of the 3 knights .......

Thus the formula is = no of ways the 1st position can be filled *no of ways the 2nd position can be filled *no of ways the 3rd position can be filled * no of ways the 4th position can be filled *no of ways the 5th position can be filled = 5*4*3*2*1 = 120

But in circular permutation for distinct objects one has to elliminate all the repeations i.e; we must count all the possibilities only once but we can take both clockwise and anticlockwise as seperate arrangements

Let X,Y,Z,P and Q be the knights

........x
q.............y
p............ z

Here , x y z p q is a possible arrangement (clockwise) and x q p z y is also a possible arrangement (anticlockwise)
But y z p q x or z p q x y or q p z y x etc.. are not distinct arrangements.
Simillarly for ,
......x
p........... z
q........... y

and others

Thus we have the formula (n-1)!, which elliminates all repeatitions.

But for not distinct objects you have only one possible combination i,e; either clockwise or anticlockwise e.g. pearls in a necklace.

Thus we have the formula (n-1)!/2

Hope this helps

Last edited by manojmakkatt on 22 Nov 2009, 07:13, edited 1 time in total.
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Re: M01 #24 [#permalink]

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20 Nov 2009, 02:20
Hi,
This question has already been discussed here:
m01-q24-78548.html

Please search this thread for the questions that were already discussed:
gmat-club-tests-master-threads-all-tests-78599.html

If you can't find the discussion you need in the thread above, please use Forum Search or Custom Google Search available at the top of every page.

Please do not create duplicate topics. Thank you for cooperation .
lagomez wrote:
Can someone give me an explanation on how to solve these probability in a circle type problems?

The formula is (N-1)! but why?

If 5 noble knights are to be seated at a round table, then how many different ways can they be seated?

120
96
60
35
24

[Reveal] Spoiler:
E

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VP
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Re: M01 #24 [#permalink]

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20 Nov 2009, 09:17
manojmakkatt wrote:
Now if we were to arrange them on a bench then we would use following formula n! i.e; 1st position can be taken by any 1 of the 5 knights, 2nd position can be taken by only 1 of the 4 knights because 1 knight has already taken the 1st position, simillarly 3rd position can be taken by any 1 of the 3 knights .......

Thus the formula is = no of ways the 1st position can be filled *no of ways the 2nd position can be filled *no of ways the 3rd position can be filled * no of ways the 4th position can be filled *no of ways the 5th position can be filled = 5*4*3*2*1 = 120

But in circular permutation for distinct objects one has to elliminate all the repeations i.e; we must count all the possibilities only once but we can take both clockwise and anticlockwise as seperate arrangements

Let X,Y,Z,P and Q be the knights

........x
q.............y
p............ z

Here , x y z p q is a possible arrangement (clockwise) and x q p z y is also a possible arrangement (anticlockwise)
But y z p q x or z p q x y or q p z y x etc.. are not distinct arrangements.
Simillarly for ,
......x
p........... z
q........... y

and others

Thus we have the formula (n-1)!, which elliminates all repeatitions.

But for not distinct objects you have only one possible combination i,e; either clockwise or anticlockwise e.g. pearls in a necklace.

Thus we have the formula (n-1)!/2

Hope this helps

thanks for the explanation
Re: M01 #24   [#permalink] 20 Nov 2009, 09:17
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# M01 #24

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