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Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]

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12 Jul 2011, 02:11

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Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

Re: How many different arrangements of finishes ? [#permalink]

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12 Jul 2011, 02:14

Alchemist1320 wrote:

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

(A) 720 (B) 360 (C) 120 (D) 24 (E) 21

no hope finishing before goldenrod= 1/2 of the times therefore the different arrangements are 1/2* 6! = 360 = B

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Re: How many different arrangements of finishes ? [#permalink]

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19 Jul 2011, 18:26

If anyone has the time, could you please elaborate on this? Why divide by 2? If the Goldenrod is always head, aren't there more than 1/2 ways for him to be ahead?

Re: How many different arrangements of finishes ? [#permalink]

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19 Jul 2011, 21:25

sudhir18n wrote:

Alchemist1320 wrote:

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

(A) 720 (B) 360 (C) 120 (D) 24 (E) 21

no hope finishing before goldenrod= 1/2 of the times therefore the different arrangements are 1/2* 6! = 360 = B

I don't get this intuition. Please kindly explain :D
_________________

There are 3 contestants A, B and C. In how many different ways can they finish a race? The following arrangements are possible A B C A C B B A C B C A C A B C B A A total of 3! = 6 arrangements. The first position is occupied by the contestant whose name is written first i.e. A B C implies A stand first, B second and C third. In how many of these is A ahead of B? We count and get 3 (A B C, A C B and C A B) In how many of these is B ahead of A? We count and get 3 again (B A C, B C A, C B A) The question is that out of 6 arrangements why is it that in half A is ahead and in the other half, B is ahead? Because the arrangements are symmetrical. Each element has the same status. Since we are taking into account all arrangements, if half of them are partial toward A, other half have to be partial toward B. There is no difference between A and B.

So if we have 6 contestants and two of them are Goldenrod and Nohope, out of a total of 6! = 720 arrangements, in 360 Goldenrod will be ahead of Nohope and in the other 360, Nohope will be ahead of Goldenrod.
_________________

Re: How many different arrangements of finishes ? [#permalink]

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19 Jul 2011, 21:57

Mahtab wrote:

sudhir18n wrote:

Alchemist1320 wrote:

Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?

(A) 720 (B) 360 (C) 120 (D) 24 (E) 21

no hope finishing before goldenrod= 1/2 of the times therefore the different arrangements are 1/2* 6! = 360 = B

I don't get this intuition. Please kindly explain :D

Hi Mahtab, Not sure why you thought this was intuition? This is actually how the symmetry works, Try any random alphabets , XYZ .. Chances of X ahead of Y will always be 1/2 . This will be true for rest of the alphabets.

Re: How many different arrangements of finishes ? [#permalink]

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20 Jul 2011, 21:06

VeritasPrepKarishma wrote:

Let's work on a simplistic example first:

There are 3 contestants A, B and C. In how many different ways can they finish a race? The following arrangements are possible A B C A C B B A C B C A C A B C B A A total of 3! = 6 arrangements. The first position is occupied by the contestant whose name is written first i.e. A B C implies A stand first, B second and C third. In how many of these is A ahead of B? We count and get 3 (A B C, A C B and C A B) In how many of these is B ahead of A? We count and get 3 again (B A C, B C A, C B A) The question is that out of 6 arrangements why is it that in half A is ahead and in the other half, B is ahead? Because the arrangements are symmetrical. Each element has the same status. Since we are taking into account all arrangements, if half of them are partial toward A, other half have to be partial toward B. There is no difference between A and B.

So if we have 6 contestants and two of them are Goldenrod and Nohope, out of a total of 6! = 720 arrangements, in 360 Goldenrod will be ahead of Nohope and in the other 360, Nohope will be ahead of Goldenrod.

Got the same solution. Just to add upon Karishma's post. PS - Hope she does not mind;) When will the arrangements not be symmetrical ?

This will happen when one letter is repeated more than other. In other words - there is higher weightage for one element - So there is more partiality and symmetry is not there... You can try with an example say AABC and see how many times A comes before B and B comes before A.

Sign off for the night on a philosophical note - Nature does not love symmetry but we humans love symmetry - All it takes for a person to look beautiful are to have his eyes, nose, ears, etc in symmetry
_________________

Labor cost for typing this post >= Labor cost for pushing the Kudos Button http://gmatclub.com/forum/kudos-what-are-they-and-why-we-have-them-94812.html

Re: How many different arrangements of finishes ? [#permalink]

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20 Jul 2011, 21:23

VeritasPrepKarishma wrote:

Let's work on a simplistic example first:

There are 3 contestants A, B and C. In how many different ways can they finish a race? The following arrangements are possible A B C A C B B A C B C A C A B C B A A total of 3! = 6 arrangements. The first position is occupied by the contestant whose name is written first i.e. A B C implies A stand first, B second and C third. In how many of these is A ahead of B? We count and get 3 (A B C, A C B and C A B) In how many of these is B ahead of A? We count and get 3 again (B A C, B C A, C B A) The question is that out of 6 arrangements why is it that in half A is ahead and in the other half, B is ahead? Because the arrangements are symmetrical. Each element has the same status. Since we are taking into account all arrangements, if half of them are partial toward A, other half have to be partial toward B. There is no difference between A and B.

So if we have 6 contestants and two of them are Goldenrod and Nohope, out of a total of 6! = 720 arrangements, in 360 Goldenrod will be ahead of Nohope and in the other 360, Nohope will be ahead of Goldenrod.

Re: How many different arrangements of finishes ? [#permalink]

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06 Aug 2011, 15:04

Alchemist1320 wrote:

Goldenrod and No Hope are in a horse race with 6 contestants.

Dont know if this question sounds little stupid, but , i want to clear it..

Shouldnt the total number of participants be considered 8. The reason is that "Goldenrod and No Hope are in a horse race with 6 contestants", my reasoning is, when i tell i am walking with 2 people, it means there are totally 3. Am i thinking wrongly or how is it ?

Goldenrod and No Hope are in a horse race with 6 contestants.

Dont know if this question sounds little stupid, but , i want to clear it..

Shouldnt the total number of participants be considered 8. The reason is that "Goldenrod and No Hope are in a horse race with 6 contestants", my reasoning is, when i tell i am walking with 2 people, it means there are totally 3. Am i thinking wrongly or how is it ?

The given statement means that Goldenrod and No Hope are a part of [highlight]a horse race with 6 contestants[/highlight] (the horse race has 6 contestants). If the question wanted to say that there are 8 contestants, it would have said, "Goldenrod and No Hope are a part of a horse race with 6 other contestants."
_________________

Re: How many different arrangements of finishes ? [#permalink]

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08 Aug 2011, 06:40

I understood why the ans is 360, but can someone explain me why my method is wrong?

My variant: there are 6 places _ _ _ _ _ _ If NH 1st there are 5 places left for another horse 1 _ _ _ _ _ (i.e. 5 variants) if NH 2nd there are 4 places possible for another horse _ 2 _ _ _ _ (i.e. 4 variants) If we continue we receive 5 * 4 * 3 * 2 *1 i.e. 5! = 120.

Re: How many different arrangements of finishes ? [#permalink]

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08 Aug 2011, 07:37

alexpiers wrote:

I understood why the ans is 360, but can someone explain me why my method is wrong?

My variant: there are 6 places _ _ _ _ _ _ If NH 1st there are 5 places left for another horse 1 _ _ _ _ _ (i.e. 5 variants) if NH 2nd there are 4 places possible for another horse _ 2 _ _ _ _ (i.e. 4 variants) If we continue we receive 5 * 4 * 3 * 2 *1 i.e. 5! = 120.

What I missed?

I am not sure whether I have found it correctly or not. I guess you have calculated the variants by taking only one horse into consideration. In this case we have to consider both the horses.

here is how i solved this problem Rule : NH always finishes ahead of G Possibilities Lets assume that NH finishes 1st. In this case, the other horses can be arranged in 5! ways - 120( Rule Satisfied) NH finishes 2nd. Apart from G other horses can be arranged in 4! ways and G can take 3 or 4 or 5 or 6 = 4!*4 = 96 NH finishes 3rd. Apart from G other horses can be arranged in 4! ways and G can take 4 or 5 or 6 = 4! * 3 = 72 NH finishes 4th. Apart from G other horses can be arranged in 4! ways and G can take 5 or 6 = 4! * 2 = 48 NH finishes 5th. Apart from G other horses can be arranged in 4! ways and G can take only 6 = 4! * 1 = 24

So the total way of arranging is 120+96+72+48+21 = 24(5+4+3+2+1) = 360

Re: How many different arrangements of finishes ? [#permalink]

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08 Aug 2011, 08:11

aravicamp wrote:

alexpiers wrote:

I understood why the ans is 360, but can someone explain me why my method is wrong?

My variant: there are 6 places _ _ _ _ _ _ If NH 1st there are 5 places left for another horse 1 _ _ _ _ _ (i.e. 5 variants) if NH 2nd there are 4 places possible for another horse _ 2 _ _ _ _ (i.e. 4 variants) If we continue we receive 5 * 4 * 3 * 2 *1 i.e. 5! = 120.

What I missed?

I am not sure whether I have found it correctly or not. I guess you have calculated the variants by taking only one horse into consideration. In this case we have to consider both the horses.

here is how i solved this problem Rule : NH always finishes ahead of G Possibilities Lets assume that NH finishes 1st. In this case, the other horses can be arranged in 5! ways - 120( Rule Satisfied) NH finishes 2nd. Apart from G other horses can be arranged in 4! ways and G can take 3 or 4 or 5 or 6 = 4!*4 = 96 NH finishes 3rd. Apart from G other horses can be arranged in 4! ways and G can take 4 or 5 or 6 = 4! * 3 = 72 NH finishes 4th. Apart from G other horses can be arranged in 4! ways and G can take 5 or 6 = 4! * 2 = 48 NH finishes 5th. Apart from G other horses can be arranged in 4! ways and G can take only 6 = 4! * 1 = 24

So the total way of arranging is 120+96+72+48+21 = 24(5+4+3+2+1) = 360

hope this helps.

Thanks! I understood my mistake: I counted only outcomes with 2 horses and forget about 4 others. That is why I have only 5*4*3*2 instead of 5!+4!+3!+2!