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

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10 Sep 2004, 01:46

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 ?

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 ?

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...

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 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.

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.

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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