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

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Goldenrod and No Hope are in a horse race with 6 contestants [#permalink] New post 10 Aug 2009, 19:40
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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?

(A) 720
(B) 360
(C) 120
(D) 24
(E) 21
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Re: GoldenRod and No Hope. [#permalink] New post 11 Aug 2009, 02:16
just think of it like this

THere are 6 possible outcomes for person A to finish the race
and 5 for person B and so on

so u should multpily them all 6X5X4X...

and u get 720
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Re: GoldenRod and No Hope. [#permalink] New post 16 Jan 2012, 05:27
combinations are symmetrical, so in half combinations goldenrod will be ahead and in exactly half it will be behind.

ans: 6!/2 = 360
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Re: GoldenRod and No Hope. [#permalink] New post 16 Jan 2012, 06:37
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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?

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

All 6 horses can finish the race in 6! way (assuming no tie).

If no tie is possible between No Hope and Goldenrod, then in half of these cases No Hope will be before Goldenrod and in half of these cases after (not necessarily right before or right after). How else? So, there are 6!/2=360 different arrangements of finishes where No Hope always finishes before Goldenrod.

Answer: B.

Similar question: six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
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Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink] New post 22 May 2013, 02:59
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Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink] New post 23 May 2013, 06:48
I have solved this in below fashion

Attachment:
gamtQ2.png
gamtQ2.png [ 2.97 KiB | Viewed 1609 times ]


So, When 'N' finishes first then there are 5 places where G can take (i.e 5 ways)
When 'N' finishes second then there are 4 places where G can take (i.e 4 ways)
When 'N' finishes third then there are 3 places where G can take (i.e 3 ways)
...

No of ways is 5*4*3*2 = 120.

please explain flaw in my explanation.

Thanks in Advance
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Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink] New post 23 May 2013, 07:39
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vishnuvardhan777 wrote:
I have solved this in below fashion

When 'N' finishes second then there are 4 places where G can take (i.e 4 ways)
When 'N' finishes third then there are 3 places where G can take (i.e 3 ways)
...

No of ways is 5*4*3*2 = 120.

please explain flaw in my explanation.

Thanks in Advance


The most effective way to solve this is Bunuel's. However if we wanna take your approach more calculus are needed:

first of all 6 spots to fill _ _ _ _ _ _
If 'N' finishes first then the other 5 spots can be filled in 5! ways N 5 4 3 2 1

And now it gets complicated...
If 'N' second then the other 5 spots can be filled in 4*4! ways 4 N 4 3 2 1. Why this?
N is in second position and there are 5 horses left. Of those only 4 can occupy the first position (every horse EXCEPT g), so write 4 on the first line.
We have 4 slots left and 4 horses => 4 3 2 1 for the remaining spots

With the same method if N finished 3rd, we get 4*3*3! ways 4 3 N 3 2 1
N on the third slot, 5 horses left. The first spot can be occupied by 4 horses (every horse EXCEPT g), the second can be occupied by three horses (every horse EXCEPT g and the previusly chosen one); then for the others spots we have 3 horses => 3! ways.
if N finished 4th, we get 4*3*2*2! ways 4 3 2 N 2 1
if N finished 5th, we get 4*3*2*1 ways 4 3 2 1 N 1

Sum them up and you get 360. This approach however is really long

Hope it's clear
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Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink] New post 23 May 2013, 07:54
I will also apply the brunel specified method.

I just wanted to know where I went wrong and correct my understanding.
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Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink] New post 23 May 2013, 08:41
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Expert's post
vishnuvardhan777 wrote:
I have solved this in below fashion

Attachment:
gamtQ2.png


So, When 'N' finishes first then there are 5 places where G can take (i.e 5 ways)
When 'N' finishes second then there are 4 places where G can take (i.e 4 ways)
When 'N' finishes third then there are 3 places where G can take (i.e 3 ways)
...

No of ways is 5*4*3*2 = 120.

please explain flaw in my explanation.

Thanks in Advance


If N is on the first place G can take ANY of the remaining 5 places and the remaining 4 horses can be arranged in 4! number of ways: 5*4!=120;
If N is on the second place G can take 4 places and the remaining 4 horses can be arranged in 4! number of ways: 4*4!=96;
If N is on the third place G can take 3 places and the remaining 4 horses can be arranged in 4! number of ways: 3*4!=72;
If N is on the fourth place G can take 2 places and the remaining 4 horses can be arranged in 4! number of ways: 2*4!=48;
If N is on the fifth place G can take only 1 place and the remaining 4 horses can be arranged in 4! number of ways: 1*4!=24;

120+96+72+48+24=360.

Answer: B.

Hope it's clear.
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Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink] New post 23 May 2013, 08:43
Expert's post
vishnuvardhan777 wrote:
I have solved this in below fashion

Attachment:
gamtQ2.png


So, When 'N' finishes first then there are 5 places where G can take (i.e 5 ways)
When 'N' finishes second then there are 4 places where G can take (i.e 4 ways)
When 'N' finishes third then there are 3 places where G can take (i.e 3 ways)
...

No of ways is 5*4*3*2 = 120.

please explain flaw in my explanation.

Thanks in Advance


Questions about the same concept to practice:
susan-john-daisy-tim-matt-and-kim-need-to-be-seated-in-130743.html
meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
in-how-many-different-ways-can-the-letters-a-a-b-91460.html
mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html

Hope it helps.
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COLLECTION OF QUESTIONS:
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Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink] New post 08 Sep 2014, 07:40
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Re: Goldenrod and No Hope are in a horse race with 6 contestants   [#permalink] 08 Sep 2014, 07:40
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