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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 Sep 2004, 00:46
Goldenrod and No Hope are in a horse race with 6 contestants . How many diffirent 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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 [#permalink] New post 10 Sep 2004, 02:19
is the answer B ?

lemme knw if it is rght ...then i will post the explanation
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 [#permalink] New post 10 Sep 2004, 03:16
I am not clear does it mean there are 8 contestans in toal or 6?
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 [#permalink] New post 10 Sep 2004, 03:26
if it is 6 then 6!/2
360
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 [#permalink] New post 10 Sep 2004, 06:33
Agree

6! various rankings and Prob (A<B) = Prob (B<A) = 1/2
so 6!/2
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Re: Permutations - Horses [#permalink] New post 10 Sep 2004, 06:54
I am getting it as B.
I solved it in a round about way.

Let the horses be N and G and N > G.
When N occupies the first spot, G can any of the 5 spots, Remaining 4 can be in any order. => 5*4! = 120
When N occupies the second spot, 4*4! = 96

Thus the total number of ways = 5*4! + 4*4! + 3*4! + 2*4! + 1*4!
= 15*4! = 15*24 = 360.

But, twixt solved it in the best way I believe.


rahul wrote:
Goldenrod and No Hope are in a horse race with 6 contestants . How many diffirent 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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Awaiting response,

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 [#permalink] New post 10 Sep 2004, 12:03
agree with B 360. Easiest way is 6! * 1/2 = 360
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 [#permalink] New post 10 Sep 2004, 12:10
Ok, I did it though but how do you guys say there are 6 or 8 contestants?
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 [#permalink] New post 10 Sep 2004, 14:48
saurya_s wrote:
Ok, I did it though but how do you guys say there are 6 or 8 contestants?
S

The question seemed to ask for 6 contestants total. Anyhow, with 8 contestants, the answer is not in the choices given.
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 [#permalink] New post 10 Sep 2004, 16:34
C. 120

NGXXXX = 4!
XNGXXX = 4!
XXNGXX = 4!
XXXNGX = 4!
XXXXNG = 4!

Therefore,

5*4! = 120

what is OA?
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 [#permalink] New post 10 Sep 2004, 16:55
viengsai wrote:
C. 120

NGXXXX = 4!
XNGXXX = 4!
XXNGXX = 4!
XXXNGX = 4!
XXXXNG = 4!

Therefore,

5*4! = 120

what is OA?

How about NXGXX - NXXGX - NXXXG - XNXGX ... and so on? :wink:
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 [#permalink] New post 10 Sep 2004, 17:19
good catch.

With correction it should be:

(6c4)*4! = 360
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Permutations---Horses [#permalink] New post 13 Sep 2004, 10:29
Guys, i don't understand why you people are dividing it by 2.

Anybody ????????????? Paul, Saurya ,Rahul
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 [#permalink] New post 13 Sep 2004, 10:53
there is a simpler way, though it won't help you understand a different problem if encountered...

the choices are 720, 360, 120, 24, and 21

if order of finish didn't matter, then the answer would be 6!, or 720, but you know there will be fewer choices, so scratch 720....

if N finished first, then the number of possible outcomes after him getting first would be 5!, which is 120, so our answer must be larger than 120 (because in can also finish 2, 3, or 4th), so scratch off the other three...

leaving 360
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 [#permalink] New post 13 Sep 2004, 12:49
1/2 is because Goldenrod, no matter where No Hope is in the race, has 1/2 probability of being in front of him. Think of it this way: If you have only those 2 in the race, what is the odd that 1, say Goldenrod, is in front of the other? The odds of that happening is just 1/2. Actually, no matter how many people will be in the race(even though there were 100 racers), there will be 1/2 a chance for Goldenrod to be ahead of No Hope given that every racer have the same chance of winning.
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Just to be Clearer [#permalink] New post 13 Sep 2004, 14:11
Just to be a bit clearer, for purposes of this question, it is not really because the Probability of N finishing ahead of B is 1/2 - just that there are in 1/2 of the total combinations N finishes ahead of B and in 1/2 of the combinations B finishes ahead of N. Thus, we need subtract the 1/2 that doesn't fit our restriction (i.e. divide by 2).

Just for conceptual purposes - on combination problems, focus on the actual # of possible combinations, not the probability of the outcomes (which theoretically could be the same, but not necessarily). Chance is not relevant here.

I know this may seem like just a picky semantic point - and I do agree that largely it is - but the semantics can sometimes throw off those who have not fully grasped it conceptually.

Hope this helps, and in no way was meant to do anything more than provide further support/clarification for paul's post.

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Permutations-Horses [#permalink] New post 13 Sep 2004, 15:40
Oh, two different concepts.

So, If N always finishes before G and O (for e.g, lets extend it to 3 horses).

If it were the case , we would divide it by 3, since there r 1/3 of the total permutations N finishes ahead of G, 1/3 of the total permutations N will be in middle and 1/3 of the total N will be in the end.

Right, so in this case the answer would be 6!/3.

Plz correct,if wrong.
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 [#permalink] New post 13 Sep 2004, 16:40
I concur with you Derek. Let's take it one step further. What are the odds that G always finishes ahead of the other 5 horses?
We then have the diagram:
G-X-X-X-X-X
When G is fixed at first position, we have 5! ways of arranging the last 5 horses. This is also equivalent to 6! * 1/6 --> 6!/6 = 5!
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 [#permalink] New post 14 Sep 2004, 07:29
Thanks Paul and mdp4d for making my concepts clear.
  [#permalink] 14 Sep 2004, 07:29
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