Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 18 Jul 2009
Posts: 12

Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
10 Aug 2009, 20:40
2
This post received KUDOS
19
This post was BOOKMARKED
Question Stats:
69% (00:55) correct 31% (01:17) wrong based on 781 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.



Intern
Joined: 26 Jul 2009
Posts: 5

Re: GoldenRod and No Hope. [#permalink]
Show Tags
11 Aug 2009, 03:16
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



Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 238

Re: GoldenRod and No Hope. [#permalink]
Show Tags
16 Jan 2012, 06:27
4
This post received KUDOS
2
This post was BOOKMARKED
combinations are symmetrical, so in half combinations goldenrod will be ahead and in exactly half it will be behind. ans: 6!/2 = 360
_________________
press +1 Kudos to appreciate posts Download Valuable Collection of Percentage Questions (PS/DS)



Math Expert
Joined: 02 Sep 2009
Posts: 45429

Re: GoldenRod and No Hope. [#permalink]
Show Tags
16 Jan 2012, 07:37
4
This post received KUDOS
Expert's post
5
This post was BOOKMARKED



Math Expert
Joined: 02 Sep 2009
Posts: 45429

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
22 May 2013, 03:59



Intern
Joined: 03 May 2013
Posts: 4
Location: India
Concentration: Economics, Marketing
GMAT Date: 07312013
WE: Information Technology (Computer Software)

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
23 May 2013, 07:48
1
This post was BOOKMARKED
I have solved this in below fashion Attachment:
gamtQ2.png [ 2.97 KiB  Viewed 7986 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



VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1101
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
23 May 2013, 08:39
6
This post received KUDOS
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 1And 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 1N 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 1if N finished 5th, we get \(4*3*2*1\) ways 4 3 2 1 N 1Sum them up and you get 360. This approach however is really long Hope it's clear
_________________
It is beyond a doubt that all our knowledge that begins with experience.
Kant , Critique of Pure Reason Tips and tricks: Inequalities , Mixture  Review: MGMAT workshop Strategy: SmartGMAT v1.0  Questions: Verbal challenge SC III CR New SC set out !! , My QuantRules for Posting in the Verbal Forum  Rules for Posting in the Quant Forum[/size][/color][/b]



Intern
Joined: 03 May 2013
Posts: 4
Location: India
Concentration: Economics, Marketing
GMAT Date: 07312013
WE: Information Technology (Computer Software)

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
23 May 2013, 08:54
I will also apply the brunel specified method.
I just wanted to know where I went wrong and correct my understanding.



Math Expert
Joined: 02 Sep 2009
Posts: 45429

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
23 May 2013, 09:41
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 45429

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
23 May 2013, 09:43



Current Student
Status: The Final Countdown
Joined: 07 Mar 2013
Posts: 288
Concentration: Technology, General Management
GPA: 3.84
WE: Information Technology (Computer Software)

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
29 Nov 2014, 07:07
@Bunuel,I used the method you explained later but i want to use the first method(All cases by 2) next time..How do i come to know in which questions i can use this approach..that in half cases one thing must have happened and in the other half cases the other thing?



Intern
Joined: 04 Mar 2014
Posts: 10
Concentration: Marketing, General Management

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
29 Nov 2014, 09:57
Bunuel wrote: 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: sixmobstershavearrivedatthetheaterforthepremiereofthe126151.html Here's my thinking: order (from 2nd8th place) of 6 horses does't matter. So we have 6!=720 choices 720 choices include 2 cases: No Hope comes first and Goldenrod comes first. Final answer is 720/2=360. Please correct me if there is anything wrong. Thanks



Intern
Joined: 06 Feb 2016
Posts: 7
Location: United Arab Emirates
Concentration: General Management, Leadership
GPA: 3.2

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
27 Mar 2016, 12:20
Hi Veterans, I have a query, Since Glen and No Hope are always fixed wr to each other, cant we just combine them into an entity and thus have 5 entities left with us which can be arranged in 5! ways ? However, this gives wrong answer. Appreciate a quick reply. Thanks a ton!



Math Expert
Joined: 02 Sep 2009
Posts: 45429

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
27 Mar 2016, 12:41



Intern
Joined: 06 Feb 2016
Posts: 7
Location: United Arab Emirates
Concentration: General Management, Leadership
GPA: 3.2

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
27 Mar 2016, 12:46
Thanks Bunuel, clear as a crystal! Bunuel wrote: mikeonbike wrote: Hi Veterans, I have a query, Since Glen and No Hope are always fixed wr to each other, cant we just combine them into an entity and thus have 5 entities left with us which can be arranged in 5! ways ? However, this gives wrong answer. Appreciate a quick reply. Thanks a ton! No Hope always finishes before Goldenrod does NOT mean that No Hope always finishes right before Goldenrod, there might be some other contestants between them.



Intern
Joined: 30 May 2016
Posts: 2

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
10 Jun 2016, 01:49
Hi Veterans,
I have a very embarrassing doubt,
The question stem reads "Goldenrod and No Hope are in a horse race with 6 contestants" doesn't that means we have 8 horses in the race ? (By calculating 6! and dividing it by 2 will only give stats about the 6 horses !!)



Senior Manager
Joined: 23 Apr 2015
Posts: 323
Location: United States
Concentration: General Management, International Business
WE: Engineering (Consulting)

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
24 Aug 2016, 13:05
MohitRulz wrote: 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 There are 6 horses and hence 6! outcomes are possible and in that 1/2 the time No Hope will be ahead of Goldenrod and 1/2 the times lag. So it's \(\frac{6!}{2}\) = 360 (B)



Senior Manager
Joined: 04 Oct 2015
Posts: 349
Location: Viet Nam
Concentration: Finance, Economics
GPA: 3.56

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
21 Aug 2017, 18:35
Hi Bunuel! I wonder if the stem is ambiguous because it only state different arrangements of finishes, i think cases in which all horses finish at the same time, or 2 horses finish at the same time, or 3 horses finish at the same time are possible. In solutions above, they readily assume that all 6 horses finish at 6 different times. How do you think?
_________________
Do not pray for an easy life, pray for the strength to endure a difficult one  Bruce Lee



Manager
Joined: 30 Mar 2017
Posts: 101

Re: Goldenrod and No Hope are in a horse race with 6 contestants [#permalink]
Show Tags
30 Dec 2017, 18:36
Hi Bunuel, how would you go about solving this if a tie was possible? Let's say if 2 other horses tied. Is this too complicated for the GMAT?




Re: Goldenrod and No Hope are in a horse race with 6 contestants
[#permalink]
30 Dec 2017, 18:36






