Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 May 2015, 21:46

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:
Manager
Joined: 18 Jun 2004
Posts: 105
Location: san jose , CA
Followers: 1

Kudos [?]: 29 [0], given: 0

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

---- Hero never chooses Destiny
Destiny chooses Him ......

Manager
Joined: 28 Jul 2004
Posts: 54
Followers: 1

Kudos [?]: 0 [0], given: 0

lemme knw if it is rght ...then i will post the explanation
_________________

Jim

Director
Joined: 08 Jul 2004
Posts: 604
Followers: 2

Kudos [?]: 82 [0], given: 0

I am not clear does it mean there are 8 contestans in toal or 6?
S
Director
Joined: 08 Jul 2004
Posts: 604
Followers: 2

Kudos [?]: 82 [0], given: 0

if it is 6 then 6!/2
360
Director
Joined: 31 Aug 2004
Posts: 606
Followers: 3

Kudos [?]: 43 [0], given: 0

Agree

6! various rankings and Prob (A<B) = Prob (B<A) = 1/2
so 6!/2
Senior Manager
Joined: 22 Jun 2004
Posts: 394
Location: Bangalore, India
Followers: 1

Kudos [?]: 2 [0], given: 0

Re: Permutations - Horses [#permalink]  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

_________________

Awaiting response,

Thnx & Rgds,
Chandra

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 27

Kudos [?]: 217 [0], given: 0

agree with B 360. Easiest way is 6! * 1/2 = 360
_________________

Best Regards,

Paul

Director
Joined: 08 Jul 2004
Posts: 604
Followers: 2

Kudos [?]: 82 [0], given: 0

Ok, I did it though but how do you guys say there are 6 or 8 contestants?
S
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 27

Kudos [?]: 217 [0], given: 0

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.
_________________

Best Regards,

Paul

Intern
Joined: 15 Jul 2004
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 0

C. 120

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

Therefore,

5*4! = 120

what is OA?
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 27

Kudos [?]: 217 [0], given: 0

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?
_________________

Best Regards,

Paul

Intern
Joined: 15 Jul 2004
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 0

good catch.

With correction it should be:

(6c4)*4! = 360
Intern
Joined: 09 Sep 2004
Posts: 16
Followers: 0

Kudos [?]: 0 [0], given: 0

Permutations---Horses [#permalink]  13 Sep 2004, 10:29
Guys, i don't understand why you people are dividing it by 2.

Anybody ????????????? Paul, Saurya ,Rahul
Intern
Joined: 03 Aug 2004
Posts: 26
Followers: 0

Kudos [?]: 0 [0], given: 0

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
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 27

Kudos [?]: 217 [0], given: 0

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.
_________________

Best Regards,

Paul

Intern
Joined: 29 Aug 2004
Posts: 4
Followers: 0

Kudos [?]: 0 [0], given: 0

Just to be Clearer [#permalink]  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.

Cheers.
Intern
Joined: 09 Sep 2004
Posts: 16
Followers: 0

Kudos [?]: 0 [0], given: 0

Permutations-Horses [#permalink]  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.
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 27

Kudos [?]: 217 [0], given: 0

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!
_________________

Best Regards,

Paul

Intern
Joined: 09 Sep 2004
Posts: 16
Followers: 0

Kudos [?]: 0 [0], given: 0

Thanks Paul and mdp4d for making my concepts clear.
Similar topics Replies Last post
Similar
Topics:
4 In a horse race, horse A runs clockwise 1 08 May 2014, 08:40
Studies on thoroughbred racing displayed that if show-horses 4 02 Mar 2012, 06:04
2 Goldenrod and No Hope are in a horse race with 6 contestants 16 12 Jul 2011, 01:11
16 Goldenrod and No Hope are in a horse race with 6 contestants 12 10 Aug 2009, 19:40
Beyond the Horse Race 8 24 Aug 2007, 20:50
Display posts from previous: Sort by