BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

permutation, a good official question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 1404
Joined: Tue May 20, 2008 6:55 pmFollowed by: 2 members
Thanked: 18 times
by tanviet » Fri Apr 17, 2009 3:57 am
Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob

24
30
60
90
120

pls, help with this
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

by bluementor » Fri Apr 17, 2009 4:14 am
There are 5! = 120 different ways to finish the race.

Out of these 120 ways, half of them will be exact mirror opposites of the other half. For example:

Lets say the 5 participants are X, Y, Z, Meg and Bob. So one of the possibilites could be:

X, Y, Z, Meg, Bob (with Bob finishing last)

an opposite of this sequence would be:

Bob, Meg, Z, Y, X (B finishes first, ahead of Meg).

So in other words, you will have opposite pairs like the above in the 120 ways (so you will have 60 pairs). Since only one possibility in each pair will have Meg finishing ahead of Bob, the answer is 60 different ways.

Choose C.

-BM-
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 322
Joined: Fri Mar 27, 2009 3:56 pm
Thanked: 24 times
GMAT Score:710

by mike22629 » Fri Apr 17, 2009 6:10 am
(5!/2) = 60
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1404
Joined: Tue May 20, 2008 6:55 pmFollowed by: 2 members
Thanked: 18 times

by tanviet » Mon Apr 20, 2009 1:34 am
still not understand, pls, explain. best regards.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

by bluementor » Mon Apr 20, 2009 2:27 am
duongthang wrote:still not understand, pls, explain. best regards.
Lets reduce the problem to only 3 people: Bob, Meg and X.

The number of ways the race can finish = 3! = 3x2x1 = 6

If you try listing the 6 different ways, you will get:

(1) BMX (Bob finishes first, Meg second, and X third)
(2) BXM
(3) XBM
(4) MBX
(5) MXB
(6) XMB

See that (1) and (6) are opposites to each other, in terms of arrangement. In (1) you have BMX, and if you reverse the sequence, you get (6) XMB. Between these two only (6) has the combination you want, i.e. Meg finishes before Bob.

The same can be done for the (2) and (5) pair, and the (3) and (4) pair. From each pair, only one combination will satisfy that Meg finishes before Bob. So in total here, you have 6 combinations, of which only 3 satisfies the criteria.

So if you generalize this further, if there are n participants in the race, then the number of ways in which Meg finishes before Bob will be:

n!/2 … this is what mike22629 used to solve this problem.

So for this problem, there are 5 participants in total. So the number of ways Meg can finish before Bob is:

5!/2 = 120/2 = 60.

Hope this helps.

-BM-
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1404
Joined: Tue May 20, 2008 6:55 pmFollowed by: 2 members
Thanked: 18 times

by tanviet » Mon Apr 20, 2009 3:57 am
thank for great explanation, friend.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members
Thanked: 82 times
GMAT Score:720

by maihuna » Tue Apr 21, 2009 8:57 am
I will prefer it positional way:

X X X B M : !4 = 24
X X X M Y : 3*3*2 = 18
X X M Y Z : 2*!3 = 12
B M X X X: !3 = 6

total = 60

prefer the positional one as it will be helpful for wider such questions rather than the pattern which will be limited in repeat and suitability.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 8212
Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members
Location: Los Angeles, CA
Thanked: 43 times

by Scott@TargetTestPrep » Wed Jan 03, 2018 7:32 am
tanviet wrote:Meg and Bob are among the 5 participants in a cycling race. If each participant finishes the race and no two participants finish at the same time, in how many different possible orders can the participants finish the race so that Meg finishes ahead of Bob

24
30
60
90
120
We can create the following equation:

total number of ways to finish the race = number of ways Meg finishes ahead of Bob + number of ways Meg does not finish ahead of Bob

Since the total number of ways to complete the race is 5! = 120, and since there are an equal number of ways for Meg to finish ahead Bob as there are for her not to finish ahead of Bob, Meg can finish ahead of Bob in 60 ways.

Alternate Solution:

If Meg comes in the first, then Bob can finish the race in any of the remaining positions, so there are 4! = 24 ways this could happen.

If Meg comes in the second, there are 3 choices for the first spot (since Bob can't finish ahead of Meg) and the last 3 spots can be filled in 3! ways; so there are 3 x 3! = 3 x 6 = 18 ways this could happen.

If Meg comes in the third, there are 3 choices for the first spot, 2 choices for the second spot, 2 choices for the fourth spot, and 1 choice for the last spot; so, this could happen in 3 x 2 x 2 = 12 ways.

Finally, if Meg comes in the fourth, then Bob must finish last and the first 3 spots can be filled in 3! = 6 ways.

Note that Meg cannot finish the race in the last position because then Bob will have finished the race ahead of Meg.

In total, there are 24 + 18 + 12 + 6 = 60 ways Meg can finish the race ahead of Bob.

Answer: C

Scott Woodbury-Stewart
Founder and CEO
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews

ImageImage
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1