• NEW! FREE Beat The GMAT Quizzes
    NEW! FREE Beat The GMAT Quizzes
    NEW! FREE Beat The GMAT Quizzes
    Hundreds of Questions Highly Detailed Reporting Expert Explanations TAKE A FREE GMAT QUIZ
  • 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 Prep
    VERITAS PRACTICE GMAT EXAMS
    Earn 10 Points Per Post
    Earn 10 Points Per Thanks
    Earn 10 Points Per Upvote
    REDEEM NOW

Arabian Horses - Good One!

This topic has 9 expert replies and 8 member replies
Goto page
  • 1,
  • 2
Next

Arabian Horses - Good One!

Post
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

  • +1 Upvote Post
  • Quote
  • Flag

GMAT/MBA Expert

Post
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 – Creator of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for our free Question of the Day emails
And check out all of our free resources

  • +1 Upvote Post
  • Quote
  • Flag
GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!
Master | Next Rank: 500 Posts Default Avatar
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
Post
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

  • +1 Upvote Post
  • Quote
  • Flag
Senior | Next Rank: 100 Posts
Joined
13 Jun 2011
Posted:
64 messages
Upvotes:
5
Target GMAT Score:
750+
Facebook Logo
Post



Last edited by tabsang on Wed Dec 12, 2012 11:52 am; edited 1 time in total

  • +1 Upvote Post
  • Quote
  • Flag
Senior | Next Rank: 100 Posts
Joined
13 Jun 2011
Posted:
64 messages
Upvotes:
5
Target GMAT Score:
750+
Facebook Logo
Post
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. Sad

  • +1 Upvote Post
  • Quote
  • Flag

GMAT/MBA Expert

Post
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 – Creator of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for our free Question of the Day emails
And check out all of our free resources

  • +1 Upvote Post
  • Quote
  • Flag
GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!

GMAT/MBA Expert

Post
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. Sad
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 – Creator of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for our free Question of the Day emails
And check out all of our free resources

  • +1 Upvote Post
  • Quote
  • Flag
GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’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+
Facebook Logo
Post
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. Sad
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!!! Smile Smile
(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 Smile Smile

Cheers,
Taz

  • +1 Upvote Post
  • Quote
  • Flag

GMAT/MBA Expert

Post
tabsang wrote:
Awesome!! Fantastic!! Thanks Brent, you rock!!! Smile Smile
(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 Smile Smile

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 – Creator of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for our free Question of the Day emails
And check out all of our free resources

  • +1 Upvote Post
  • Quote
  • Flag
GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’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+
Facebook Logo
Post
Brent@GMATPrepNow wrote:
tabsang wrote:
Awesome!! Fantastic!! Thanks Brent, you rock!!! Smile Smile
(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 Smile Smile

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 Smile Smile Smile
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 Smile Smile

Cheers,
Taz

  • +1 Upvote Post
  • Quote
  • Flag
Newbie | Next Rank: 10 Posts Default Avatar
Joined
19 Jul 2011
Posted:
9 messages
Post
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

  • +1 Upvote Post
  • Quote
  • Flag

GMAT/MBA Expert

Post
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 – Creator of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Sign up for our free Question of the Day emails
And check out all of our free resources

  • +1 Upvote Post
  • Quote
  • Flag
GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months!
Junior | Next Rank: 30 Posts Default Avatar
Joined
27 Feb 2015
Posted:
25 messages
Upvotes:
1
Post
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?

  • +1 Upvote Post
  • Quote
  • Flag
Junior | Next Rank: 30 Posts Default Avatar
Joined
27 Feb 2015
Posted:
25 messages
Upvotes:
1
Post
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.

  • +1 Upvote Post
  • Quote
  • Flag

GMAT/MBA Expert

GMAT Instructor Default Avatar
Joined
12 Sep 2012
Posted:
2635 messages
Followed by:
115 members
Upvotes:
625
Target GMAT Score:
V51
GMAT Score:
780
Post
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!

  • +1 Upvote Post
  • Quote
  • Flag
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!
  • Target Test Prep
    5-Day Free Trial
    5-day free, full-access trial TTP Quant

    Available with Beat the GMAT members only code

    MORE DETAILS
    Target Test Prep
  • Veritas Prep
    Free Veritas GMAT Class
    Experience Lesson 1 Live Free

    Available with Beat the GMAT members only code

    MORE DETAILS
    Veritas Prep
  • e-gmat Exclusive Offer
    Get 300+ Practice Questions
    25 Video lessons and 6 Webinars for FREE

    Available with Beat the GMAT members only code

    MORE DETAILS
    e-gmat Exclusive Offer
  • Magoosh
    Magoosh
    Study with Magoosh GMAT prep

    Available with Beat the GMAT members only code

    MORE DETAILS
    Magoosh
  • EMPOWERgmat Slider
    1 Hour Free
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    EMPOWERgmat Slider
  • Varsity Tutors
    Award-winning private GMAT tutoring
    Register now and save up to $200

    Available with Beat the GMAT members only code

    MORE DETAILS
    Varsity Tutors
  • PrepScholar GMAT
    5 Day FREE Trial
    Study Smarter, Not Harder

    Available with Beat the GMAT members only code

    MORE DETAILS
    PrepScholar GMAT
  • Kaplan Test Prep
    Free Practice Test & Review
    How would you score if you took the GMAT

    Available with Beat the GMAT members only code

    MORE DETAILS
    Kaplan Test Prep
  • The Princeton Review
    FREE GMAT Exam
    Know how you'd score today for $0

    Available with Beat the GMAT members only code

    MORE DETAILS
    The Princeton Review
  • Economist Test Prep
    Free Trial & Practice Exam
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    Economist Test Prep

Top First Responders*

1 Brent@GMATPrepNow 56 first replies
2 fskilnik@GMATH 55 first replies
3 GMATGuruNY 50 first replies
4 Jay@ManhattanReview 38 first replies
5 Rich.C@EMPOWERgma... 23 first replies
* Only counts replies to topics started in last 30 days
See More Top Beat The GMAT Members

Most Active Experts

1 image description Scott@TargetTestPrep

Target Test Prep

198 posts
2 image description fskilnik@GMATH

GMATH Teacher

157 posts
3 image description Brent@GMATPrepNow

GMAT Prep Now Teacher

105 posts
4 image description Max@Math Revolution

Math Revolution

83 posts
5 image description GMATGuruNY

The Princeton Review Teacher

75 posts
See More Top Beat The GMAT Experts