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How many ways in a horse race with 7 horses can Horse A get

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How many ways in a horse race with 7 horses can Horse A get [#permalink] New post 08 May 2004, 09:17
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How many ways in a horse race with 7 horses can Horse A get before Horse B, but Horse C is not adjacent to either of them?

Last edited by hallelujah1234 on 08 May 2004, 10:18, edited 1 time in total.
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 [#permalink] New post 08 May 2004, 10:05
ultra cryptic :?
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 [#permalink] New post 09 May 2004, 09:17
I believe it is 7! - ( 2 * 5! + 2 * 5! ) = 5040 - ( 480 ) = 4560
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 [#permalink] New post 12 May 2004, 06:27
anandnk wrote:
I believe it is 7! - ( 2 * 5! + 2 * 5! ) = 5040 - ( 480 ) = 4560


No, you are wrong. this number is fewer than 7!/2 = 2520, because 7! stands for all possible outcomes, 7!/2 - for outcomes with A earlier than B.

From these 2520 there are combinations for which C is adjacent to either A or B. So, the final response is smaller than 2520.
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Re: Horse Race PS-32 [#permalink] New post 12 May 2004, 06:47
hallelujah1234 wrote:
How many ways in a horse race with 7 horses can Horse A get before Horse B, but Horse C is not adjacent to either of them?


By the way, what are possible answers?

#<= 2520. What are other suggestions?

Huh, even more: for every such combination there are 4! clones... So, this # should :: 24.

Last edited by Emmanuel on 12 May 2004, 06:53, edited 2 times in total.
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 [#permalink] New post 12 May 2004, 06:50
Yeah I realize my mistake

So no. of combinations = 7!/2 = 2520

Out of this B and C will be adjacent 2*6!/2 times = 720

I believe the answer should be 2520-720 = 1800.
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 [#permalink] New post 12 May 2004, 06:53
anandnk wrote:
Yeah I realize my mistake

So no. of combinations = 7!/2 = 2520

Out of this B and C will be adjacent 2*6!/2 times = 720

I believe the answer should be 2520-720 = 1800.


I doubt if it can be done so easily. Because "adjacent" means "close to", so, it can happen that A and B are adjacent, and there is no room for C between them. C is adjacent to A when AC or CA take place. Am I right?
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 [#permalink] New post 12 May 2004, 06:58
I misread the question. The stem says horse C is not adjacent to either of them. Then no of arrangements = 3! * 5! out of this A ia ahead of B half the time
I believe it is time for halle to post the answer.

3! * 5! /2 = 360
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 [#permalink] New post 12 May 2004, 07:08
anandnk wrote:
I misread the question. The stem says horse C is not adjacent to either of them. Then no of arrangements = 3! * 5! out of this A ia ahead of B half the time
I believe it is time for halle to post the answer.

3! * 5! /2 = 360


Moreover, the answer should be such that = 24 mod 48. Because for every pattern of A,B,C there exists an inverse pattern, with only one exception of (*,*,A,C,B,*,*). Inverse = {change A & B in (*,*,B,C,A,*,*)}.
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 [#permalink] New post 12 May 2004, 11:49
my guess
(7! -(2*5! + 2*5!))/2 = 2280
{C would never be adjacent to A * B in 4540 ways) and A would be ahead of B in half of them.
---trying to think simple here--
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lets do it together....

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 [#permalink] New post 12 May 2004, 21:23
sachy wrote:
my guess
(7! -(2*5! + 2*5!))/2 = 2280
{C would never be adjacent to A * B in 4540 ways) and A would be ahead of B in half of them.
---trying to think simple here--


Your answer looks like normal, since it has 24 mod 48... And is less than 2520.
  [#permalink] 12 May 2004, 21:23
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How many ways in a horse race with 7 horses can Horse A get

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