• NEW! FREE Beat The GMAT Quizzes
Hundreds of Questions Highly Detailed Reporting Expert Explanations
• 7 CATs FREE!
If you earn 100 Forum Points

Engage in the Beat The GMAT forums to earn
100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Arabian Horses - Good One! tagged by: tabsang ##### This topic has 9 expert replies and 8 member replies Goto page • 1, • 2 ## Arabian Horses - Good One! Ten Arabian horses are split into pairs to pull one of the four distinct carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible? a) 420 b) 1260 c) 5220 d) 9450 e) 113400 OA is E ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 13046 messages Followed by: 1253 members Upvotes: 5254 GMAT Score: 770 tabsang wrote: Ten Arabian horses are split into pairs to pull one of the four distinct carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible? a) 420 b) 1260 c) 5220 d) 9450 e) 113400 OA is E Take the task of assigning the 4 carts break it into stages. Stage 1: Assign the first cart to 2 horses There are 10 horses and we must select 2. Since the order of the selected horses does not matter, we can use combinations. We can select 2 horses from 10 horses in 10C2 ways (45 ways). Aside: If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789 Stage 2: Assign the second cart to 2 horses There are 8 horses remaining and we must select 2. We can select 2 horses from 8 horses in 8C2 ways (28 ways). Stage 3: Assign the third cart to 2 horses There are 6 horses remaining and we must select 2. We can select 2 horses from 6 horses in 6C2 ways (15 ways). Stage 4: Assign the fourth cart to 2 horses There are 4 horses remaining and we must select 2. We can select 2 horses from 8 horses in 4C2 ways (6 ways). By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus assign horses to all 4 carts) in (45)(28)(15)(6) ways (= 113,400 ways) Answer = E Cheers, Brent Aside: For more information about the FCP, we have a free video on the subject: http://www.gmatprepnow.com/module/gmat-counting?id=775 _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! Master | Next Rank: 500 Posts Joined 14 Nov 2012 Posted: 131 messages Followed by: 2 members Upvotes: 39 Test Date: 01/12/2013 Target GMAT Score: 700+ GMAT Score: NA Note:- This question should be out of scope for GMAT. The solution is: - Number of ways 10 horses can be divided into 5 groups when order of the groups does not matter is: (10C2*8C2*6C2*4C2*2C2)/5! = 945 Now to assign a pair to one of 4 distinct is 5P4 = 120. Total is 945 * 120 = 113400 Senior | Next Rank: 100 Posts Joined 13 Jun 2011 Posted: 64 messages Upvotes: 5 Target GMAT Score: 750+ Last edited by tabsang on Wed Dec 12, 2012 11:52 am; edited 1 time in total Senior | Next Rank: 100 Posts Joined 13 Jun 2011 Posted: 64 messages Upvotes: 5 Target GMAT Score: 750+ Brent@GMATPrepNow wrote: tabsang wrote: Ten Arabian horses are split into pairs to pull one of the four distinct carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible? a) 420 b) 1260 c) 5220 d) 9450 e) 113400 OA is E Take the task of assigning the 4 carts break it into stages. Stage 1: Assign the first cart to 2 horses There are 10 horses and we must select 2. Since the order of the selected horses does not matter, we can use combinations. We can select 2 horses from 10 horses in 10C2 ways (45 ways). Aside: If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789 Stage 2: Assign the second cart to 2 horses There are 8 horses remaining and we must select 2. We can select 2 horses from 8 horses in 8C2 ways (28 ways). Stage 3: Assign the third cart to 2 horses There are 6 horses remaining and we must select 2. We can select 2 horses from 6 horses in 6C2 ways (15 ways). Stage 4: Assign the fourth cart to 2 horses There are 4 horses remaining and we must select 2. We can select 2 horses from 8 horses in 4C2 ways (6 ways). By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus assign horses to all 4 carts) in (45)(28)(15)(6) ways (= 113,400 ways) Answer = E Cheers, Brent Aside: For more information about the FCP, we have a free video on the subject: http://www.gmatprepnow.com/module/gmat-counting?id=775 Thanks Brent for the prompt and detailed answer. However, there's another problem. I had, infact, solved the problem using the same approach as the one you took as opposed to the formula driven one. But this approach seemingly fails if I try and apply it to this question: In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people? a) 280 b) 1,260 c) 1,680 d) 2,520 e) 3,360 OA is a. If I apply the same logic as that I used in the Arabian horses problem then the answer comes out to be c . Please help. I'm confused. ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 13046 messages Followed by: 1253 members Upvotes: 5254 GMAT Score: 770 puneetkhurana2000 wrote: Note:- This question should be out of scope for GMAT. I think this is a 700+ level question, but I don't think it's out of scope (IMHO) Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 13046 messages Followed by: 1253 members Upvotes: 5254 GMAT Score: 770 tabsang wrote: Brent@GMATPrepNow wrote: tabsang wrote: Ten Arabian horses are split into pairs to pull one of the four distinct carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible? a) 420 b) 1260 c) 5220 d) 9450 e) 113400 OA is E Take the task of assigning the 4 carts break it into stages. Stage 1: Assign the first cart to 2 horses There are 10 horses and we must select 2. Since the order of the selected horses does not matter, we can use combinations. We can select 2 horses from 10 horses in 10C2 ways (45 ways). Aside: If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789 Stage 2: Assign the second cart to 2 horses There are 8 horses remaining and we must select 2. We can select 2 horses from 8 horses in 8C2 ways (28 ways). Stage 3: Assign the third cart to 2 horses There are 6 horses remaining and we must select 2. We can select 2 horses from 6 horses in 6C2 ways (15 ways). Stage 4: Assign the fourth cart to 2 horses There are 4 horses remaining and we must select 2. We can select 2 horses from 8 horses in 4C2 ways (6 ways). By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus assign horses to all 4 carts) in (45)(28)(15)(6) ways (= 113,400 ways) Answer = E Cheers, Brent Aside: For more information about the FCP, we have a free video on the subject: http://www.gmatprepnow.com/module/gmat-counting?id=775 Thanks Brent for the prompt and detailed answer. However, there's another problem. I had, infact, solved the problem using the same approach as the one you took as opposed to the formula driven one. But this approach seemingly fails if I try and apply it to this question: In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people? a) 280 b) 1,260 c) 1,680 d) 2,520 e) 3,360 OA is a. If I apply the same logic as that I used in the Arabian horses problem then the answer comes out to be c . Please help. I'm confused. If the 3 groups are distinct (e.g., being in group A is different from being in group B) then the correct answer to this question is, indeed, C. However, if the 3 groups are not distinct, then the correct answer is A. From the wording of the question, it's difficult to determine whether the 3 groups are distinct. This ambiguity makes the question unanswerable. Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! Senior | Next Rank: 100 Posts Joined 13 Jun 2011 Posted: 64 messages Upvotes: 5 Target GMAT Score: 750+ Brent@GMATPrepNow wrote: tabsang wrote: Brent@GMATPrepNow wrote: tabsang wrote: Ten Arabian horses are split into pairs to pull one of the four distinct carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible? a) 420 b) 1260 c) 5220 d) 9450 e) 113400 OA is E Take the task of assigning the 4 carts break it into stages. Stage 1: Assign the first cart to 2 horses There are 10 horses and we must select 2. Since the order of the selected horses does not matter, we can use combinations. We can select 2 horses from 10 horses in 10C2 ways (45 ways). Aside: If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789 Stage 2: Assign the second cart to 2 horses There are 8 horses remaining and we must select 2. We can select 2 horses from 8 horses in 8C2 ways (28 ways). Stage 3: Assign the third cart to 2 horses There are 6 horses remaining and we must select 2. We can select 2 horses from 6 horses in 6C2 ways (15 ways). Stage 4: Assign the fourth cart to 2 horses There are 4 horses remaining and we must select 2. We can select 2 horses from 8 horses in 4C2 ways (6 ways). By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus assign horses to all 4 carts) in (45)(28)(15)(6) ways (= 113,400 ways) Answer = E Cheers, Brent Aside: For more information about the FCP, we have a free video on the subject: http://www.gmatprepnow.com/module/gmat-counting?id=775 Thanks Brent for the prompt and detailed answer. However, there's another problem. I had, infact, solved the problem using the same approach as the one you took as opposed to the formula driven one. But this approach seemingly fails if I try and apply it to this question: In how many different ways can a group of 9 people be divided into 3 groups, with each group containing 3 people? a) 280 b) 1,260 c) 1,680 d) 2,520 e) 3,360 OA is a. If I apply the same logic as that I used in the Arabian horses problem then the answer comes out to be c . Please help. I'm confused. If the 3 groups are distinct (e.g., being in group A is different from being in group B) then the correct answer to this question is, indeed, C. However, if the 3 groups are not distinct, then the correct answer is A. From the wording of the question, it's difficult to determine whether the 3 groups are distinct. This ambiguity makes the question unanswerable. Cheers, Brent Awesome!! Fantastic!! Thanks Brent, you rock!!! (I'm so happy that the non-formula approach holds true, albeit only if the groups are distinct) Brent, I have another quick question to bring a full closure to this problem and concept: In case the groups were not distinct, what would be the approach to solve the problem? I know of a formula that we can use when the order is not important or as in this case the groups are not distinct (for (mxn) things distributed equally in groups of m each containing n objects and the order of the groups is not important: (mn)!/[(n!^m)*m!]) Please tell me that there is a non-formula driven approach to this and it'll make my day Cheers, Taz ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 13046 messages Followed by: 1253 members Upvotes: 5254 GMAT Score: 770 tabsang wrote: Awesome!! Fantastic!! Thanks Brent, you rock!!! (I'm so happy that the non-formula approach holds true, albeit only if the groups are distinct) Brent, I have another quick question to bring a full closure to this problem and concept: In case the groups were not distinct, what would be the approach to solve the problem? I know of a formula that we can use when the order is not important or as in this case the groups are not distinct (for (mxn) things distributed equally in groups of m each containing n objects and the order of the groups is not important: (mn)!/[(n!^m)*m!]) Please tell me that there is a non-formula driven approach to this and it'll make my day Cheers, Taz I suggest that you first pretend that the n groups are distinct. Then, to handle the fact that they are not distinct, divide your earlier result by n! We divide by n! because we can take n things (groups) and arrange them in n! ways. So, for the question about 9 people, we'll first assume that the groups are distinct. When we do so, we see that there are 1680 different possibilities. Then, to account for the fact that the 3 groups are not distinct, divide 1680 by 3! to get 280 Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! Senior | Next Rank: 100 Posts Joined 13 Jun 2011 Posted: 64 messages Upvotes: 5 Target GMAT Score: 750+ Brent@GMATPrepNow wrote: tabsang wrote: Awesome!! Fantastic!! Thanks Brent, you rock!!! (I'm so happy that the non-formula approach holds true, albeit only if the groups are distinct) Brent, I have another quick question to bring a full closure to this problem and concept: In case the groups were not distinct, what would be the approach to solve the problem? I know of a formula that we can use when the order is not important or as in this case the groups are not distinct (for (mxn) things distributed equally in groups of m each containing n objects and the order of the groups is not important: (mn)!/[(n!^m)*m!]) Please tell me that there is a non-formula driven approach to this and it'll make my day Cheers, Taz I suggest that you first pretend that the n groups are distinct. Then, to handle the fact that they are not distinct, divide your earlier result by n! We divide by n! because we can take n things (groups) and arrange them in n! ways. So, for the question about 9 people, we'll first assume that the groups are distinct. When we do so, we see that there are 1680 different possibilities. Then, to account for the fact that the 3 groups are not distinct, divide 1680 by 3! to get 280 Cheers, Brent Thanks Brent. You, sir, are a life saver I had been at this for quite some time and now it rests in peace with all concepts understood and learnt. Have a great day Cheers, Taz Newbie | Next Rank: 10 Posts Joined 19 Jul 2011 Posted: 9 messages Brent@GMATPrepNow wrote: tabsang wrote: Ten Arabian horses are split into pairs to pull one of the four distinct carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible? a) 420 b) 1260 c) 5220 d) 9450 e) 113400 OA is E Take the task of assigning the 4 carts break it into stages. Stage 1: Assign the first cart to 2 horses There are 10 horses and we must select 2. Since the order of the selected horses does not matter, we can use combinations. We can select 2 horses from 10 horses in 10C2 ways (45 ways). Aside: If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789 Stage 2: Assign the second cart to 2 horses There are 8 horses remaining and we must select 2. We can select 2 horses from 8 horses in 8C2 ways (28 ways). Stage 3: Assign the third cart to 2 horses There are 6 horses remaining and we must select 2. We can select 2 horses from 6 horses in 6C2 ways (15 ways). Stage 4: Assign the fourth cart to 2 horses There are 4 horses remaining and we must select 2. We can select 2 horses from 8 horses in 4C2 ways (6 ways). By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus assign horses to all 4 carts) in (45)(28)(15)(6) ways (= 113,400 ways) Answer = E Cheers, Brent Aside: For more information about the FCP, we have a free video on the subject: http://www.gmatprepnow.com/module/gmat-counting?id=775 Hi Brent, This is a great explanation. I have a question here. If it were mentioned in the question that the carts are not 'distinct', would we have divided (45)(28)(15)(6) by 4! ? Thanks, Muhtasim Hassan ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 13046 messages Followed by: 1253 members Upvotes: 5254 GMAT Score: 770 muhtasimhassan wrote: Brent@GMATPrepNow wrote: tabsang wrote: Ten Arabian horses are split into pairs to pull one of the four distinct carts in a race. If each cart is assigned to a pair, how many different assignments of horses to carts are possible? a) 420 b) 1260 c) 5220 d) 9450 e) 113400 OA is E Take the task of assigning the 4 carts break it into stages. Stage 1: Assign the first cart to 2 horses There are 10 horses and we must select 2. Since the order of the selected horses does not matter, we can use combinations. We can select 2 horses from 10 horses in 10C2 ways (45 ways). Aside: If anyone is interested, we have a free video on calculating combinations (like 10C2) in your head: http://www.gmatprepnow.com/module/gmat-counting?id=789 Stage 2: Assign the second cart to 2 horses There are 8 horses remaining and we must select 2. We can select 2 horses from 8 horses in 8C2 ways (28 ways). Stage 3: Assign the third cart to 2 horses There are 6 horses remaining and we must select 2. We can select 2 horses from 6 horses in 6C2 ways (15 ways). Stage 4: Assign the fourth cart to 2 horses There are 4 horses remaining and we must select 2. We can select 2 horses from 8 horses in 4C2 ways (6 ways). By the Fundamental Counting Principle (FCP), we can complete all 4 stages (and thus assign horses to all 4 carts) in (45)(28)(15)(6) ways (= 113,400 ways) Answer = E Cheers, Brent Aside: For more information about the FCP, we have a free video on the subject: http://www.gmatprepnow.com/module/gmat-counting?id=775 Hi Brent, This is a great explanation. I have a question here. If it were mentioned in the question that the carts are not 'distinct', would we have divided (45)(28)(15)(6) by 4! ? Thanks, Muhtasim Hassan Yes, that is correct. Cheers, Brent _________________ Brent Hanneson â€“ Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMATâ€™s FREE 60-Day Study Guide and reach your target score in 2 months! Junior | Next Rank: 30 Posts Joined 27 Feb 2015 Posted: 25 messages Upvotes: 1 Hi Brent, If the question is rephrased as below In how many different ways can a group of 9 people be divided into 3 groups? How we will solve this? Junior | Next Rank: 30 Posts Joined 27 Feb 2015 Posted: 25 messages Upvotes: 1 Hi Brent, 144 234 126 333 522 711 531 144 can arranged in 3!/2! = 3 ways 234 can arranged in 3! = 6 ways 126 can arranged in 3! = 6 ways 333 can arranged in 1! = 1 way 522 can arranged in 3!/2! = 3 ways 711 can arranged in 3!/2! = 3 ways 531 can arranged in 3! = 6 ways Total number of ways = 28 ways. ### GMAT/MBA Expert GMAT Instructor Joined 12 Sep 2012 Posted: 2635 messages Followed by: 117 members Upvotes: 625 Target GMAT Score: V51 GMAT Score: 780 Kaustubhk wrote: Hi Brent, If the question is rephrased as below In how many different ways can a group of 9 people be divided into 3 groups? How we will solve this? That's a lot more complicated! We'd need to know whether the groups are distinct, but assuming that they're not: First, choose 3 people for the first group: (9 choose 3) Next, choose 3 people for the second group: (6 choose 3) The three people remaining form the third group, so we're almost done. The only step left is to remove all the duplicates. For instance, we could have Al, Bill, and Carl as group 1, but we could also have Al, Bill, and Carl as group 2. To eliminate the duplicate arrangements, we divide by all the ways we could arrange the groups once they're formed: 3 * 2 * 1. That leaves us with (9 choose 3) * (6 choose 3) / 3! Enroll in a Veritas Prep GMAT class completely for FREE. Wondering if a GMAT course is right for you? Attend the first class session of an actual GMAT course, either in-person or live online, and see for yourself why so many students choose to work with Veritas Prep. Find a class now! • FREE GMAT Exam Know how you'd score today for$0

Available with Beat the GMAT members only code

• Magoosh
Study with Magoosh GMAT prep

Available with Beat the GMAT members only code

• Free Veritas GMAT Class
Experience Lesson 1 Live Free

Available with Beat the GMAT members only code

• Free Trial & Practice Exam
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• 5-Day Free Trial
5-day free, full-access trial TTP Quant

Available with Beat the GMAT members only code

• Free Practice Test & Review
How would you score if you took the GMAT

Available with Beat the GMAT members only code

• Award-winning private GMAT tutoring
Register now and save up to \$200

Available with Beat the GMAT members only code

• Get 300+ Practice Questions

Available with Beat the GMAT members only code

• 1 Hour Free
BEAT THE GMAT EXCLUSIVE

Available with Beat the GMAT members only code

• 5 Day FREE Trial
Study Smarter, Not Harder

Available with Beat the GMAT members only code

### Top First Responders*

1 Ian Stewart 44 first replies
2 Jay@ManhattanReview 35 first replies
3 Brent@GMATPrepNow 34 first replies
4 Scott@TargetTestPrep 31 first replies
5 GMATGuruNY 18 first replies
* Only counts replies to topics started in last 30 days
See More Top Beat The GMAT Members

### Most Active Experts

1 Scott@TargetTestPrep

Target Test Prep

129 posts
2 Max@Math Revolution

Math Revolution

89 posts
3 Ian Stewart

GMATiX Teacher

53 posts
4 Brent@GMATPrepNow

GMAT Prep Now Teacher

50 posts
5 Jay@ManhattanReview

Manhattan Review

35 posts
See More Top Beat The GMAT Experts