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.

Combinatorics - 700+

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 208
Joined: Thu Aug 26, 2010 3:10 pmFollowed by: 2 members
Thanked: 20 times
by hja379 » Fri Jan 21, 2011 5:20 am
Can you please explain? I couldn't understand how to attack this problem.

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?

A) 6
B) 24
C) 120
D) 360
E) 720

[spoiler]Ignoring Frankie's requirement for a moment, observe that the six mobsters can be arranged 6! or 6 x 5 x 4 x 3 x 2 x 1 = 720 different ways in the concession stand line. In each of those 720 arrangements, Frankie must be either ahead of or behind Joey. Logically, since the combinations favor neither Frankie nor Joey, each would be behind the other in precisely half of the arrangements. Therefore, in order to satisfy Frankie's requirement, the six mobsters could be arranged in 720/2 = 360 different ways.

The correct answer is D.[/spoiler]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Legendary Member
Posts: 586
Joined: Tue Jan 19, 2010 4:38 amFollowed by: 5 members
Thanked: 31 times
GMAT Score:730

by rohu27 » Fri Jan 21, 2011 5:35 am
This can be attacked in a smilar way as a we do a combinations word problem where in one word needs to be after a specified one. Here F(Frankie) is placed after J(Joey). So if J is to stand 1st among all the 6, F can be placed behind him in 5 ways, rest of the 4 can be arranged in 4! ways. This is one combo.

Next if J is to be placed second among al the 6 in the queue, F can be placed in 5 ways (as F is always behind J ). Remaining 4 can be arranged in 4! ways. This is second combo.

Similarly J can be the 3rd among 6 to stand in the line. So F can be placed in 3 ways..so on.

Finally we have somethign like this:

Total combinations:

5*4!+4*4!+3*4!+2*4!+1*4! = 360.

Hope this explanation helps.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 1449
Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members
Thanked: 59 times

by fskilnik@GMATH » Fri Jan 21, 2011 6:14 am
hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?

A) 6
B) 24
C) 120
D) 360
E) 720
Hi there!

Consider A and B any two of 6 people ordered in a queue.

There are 6! possible ways of arranging all of them. Half this number has A somewhere in front of B, right?

The answer to your problem is therefore 6!/2 = 6*5*4*3 = 30*12 = 360.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Fri Jan 21, 2011 7:00 am
hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?

A) 6
B) 24
C) 120
D) 360
E) 720
There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. In every arrangement, either Frankie will be behind Joey or Joey will be behind Frankie. Isn't the probability that Frankie will be behind Joey the same as the probability that Joey will be behind Frankie? Thus:

In 1/2 * 720 = 360 of these arrangements, Frankie will be behind Joey.
In 1/2 * 720 = 360 of these arrangements, Joey will be behind Frankie.

So there are 360 ways in which Frankie can be placed behind Joey.

The correct answer is D.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 208
Joined: Thu Aug 26, 2010 3:10 pmFollowed by: 2 members
Thanked: 20 times

by hja379 » Sat Jan 22, 2011 9:04 am
Thank you all so much. Great forum... with smart people !
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 69
Joined: Sat Jul 31, 2010 9:27 pmFollowed by: 4 members
Thanked: 2 times
GMAT Score:640

by Jayanth2689 » Fri Jun 10, 2011 10:07 am
GMATGuruNY wrote:
hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?

A) 6
B) 24
C) 120
D) 360
E) 720
There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. In every arrangement, either Frankie will be behind Joey or Joey will be behind Frankie. Isn't the probability that Frankie will be behind Joey the same as the probability that Joey will be behind Frankie? Thus:

In 1/2 * 720 = 360 of these arrangements, Frankie will be behind Joey.
In 1/2 * 720 = 360 of these arrangements, Joey will be behind Frankie.

So there are 360 ways in which Frankie can be placed behind Joey.

The correct answer is D.
@ GG - is it possible to solve the same using the slot method (since restrictions are present) ?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Fri Jun 10, 2011 10:36 am
Jayanth2689 wrote:
GMATGuruNY wrote:
hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?

A) 6
B) 24
C) 120
D) 360
E) 720
There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. In every arrangement, either Frankie will be behind Joey or Joey will be behind Frankie. Isn't the probability that Frankie will be behind Joey the same as the probability that Joey will be behind Frankie? Thus:

In 1/2 * 720 = 360 of these arrangements, Frankie will be behind Joey.
In 1/2 * 720 = 360 of these arrangements, Joey will be behind Frankie.

So there are 360 ways in which Frankie can be placed behind Joey.

The correct answer is D.
@ GG - is it possible to solve the same using the slot method (since restrictions are present) ?
Joey in Slot 1:
Number of ways to arrange the other 5 mobsters = 5*4*3*2*1 = 120.

Joey in Slot 2:
Since Frankie cannot be in Slot 1, the number of choices for Slot 1 = 4.
Number of ways to arrange the 4 remaining mobsters in Slots 3, 4, 5 and 6 = 4*3*2*1 = 24.
Multiplying, we get 4*24 = 96 arrangements.

Joey in Slot 3:
Since Frankie cannot be in Slots 1 or 2, the number of ways to arrange 2 of the remaining 4 mobsters in Slots 1 and 2 = 4*3 = 12.
Number of ways to arrange the remaining 3 mobsters in Slots 4, 5 and 6 = 3*2*1 = 6.
Multiplying, we get 12*6 = 72 arrangements.

Joey in Slot 4:
Since Frankie cannot be in Slots 1, 2 or 3, the number of ways to arrange 3 of the remaining 4 mobsters in Slots 1, 2 and 3 = 4*3*2 = 24.
Number of ways to arrange the remaining 2 mobsters in Slots 5 and 6 = 2*1 = 2.
Multiplying, we get 24*2 = 48 arrangements.

Joey in Slot 5:
Since Frankie must be in Slot 6, the number of choices for Slot 6 = 1.
The number of ways to arrange the remaining 4 mobsters in Slots 1, 2, 3, and 4 = 4*3*2*1 = 24.
Multiplying, we get 1*24 = 24 arrangements.

Adding the results above, we get:
120+96+72+48+24 = 360 arrangements.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 22
Joined: Thu Feb 24, 2011 4:47 am
Thanked: 1 times

Solution approach

by sunilramu » Fri Jun 10, 2011 10:36 am
If there were no restrictions then we could have 6! = 720 ways

If we have Frankie in the first slot we will have 5! = 120 ways

But Frankie can be in other slots as well. So the only answer 120 < 360 < 720 should be the solution.

IMO D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 69
Joined: Sat Jul 31, 2010 9:27 pmFollowed by: 4 members
Thanked: 2 times
GMAT Score:640

by Jayanth2689 » Fri Jun 10, 2011 10:54 am
GMATGuruNY wrote:
Jayanth2689 wrote:
GMATGuruNY wrote:
hja379 wrote:Can you please explain? I couldn't understand how to attack this problem.

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie's requirement is satisfied?

A) 6
B) 24
C) 120
D) 360
E) 720
There are 6! = 720 total ways to arrange the 6 mobsters.

Now let's think about this. In every arrangement, either Frankie will be behind Joey or Joey will be behind Frankie. Isn't the probability that Frankie will be behind Joey the same as the probability that Joey will be behind Frankie? Thus:

In 1/2 * 720 = 360 of these arrangements, Frankie will be behind Joey.
In 1/2 * 720 = 360 of these arrangements, Joey will be behind Frankie.

So there are 360 ways in which Frankie can be placed behind Joey.

The correct answer is D.
@ GG - is it possible to solve the same using the slot method (since restrictions are present) ?
Joey in Slot 1:
Number of ways to arrange the other 5 mobsters = 5*4*3*2*1 = 120.

Joey in Slot 2:
Since Frankie cannot be in Slot 1, the number of choices for Slot 1 = 4.
Number of ways to arrange the 4 remaining mobsters in Slots 3, 4, 5 and 6 = 4*3*2*1 = 24.
Multiplying, we get 4*24 = 96 arrangements.

Joey in Slot 3:
Since Frankie cannot be in Slots 1 or 2, the number of ways to arrange 2 of the remaining 4 mobsters in Slots 1 and 2 = 4*3 = 12.
Number of ways to arrange the remaining 3 mobsters in Slots 4, 5 and 6 = 3*2*1 = 6.
Multiplying, we get 12*6 = 72 arrangements.

Joey in Slot 4:
Since Frankie cannot be in Slots 1, 2 or 3, the number of ways to arrange 3 of the remaining 4 mobsters in Slots 1, 2 and 3 = 4*3*2 = 24.
Number of ways to arrange the remaining 2 mobsters in Slots 5 and 6 = 2*1 = 2.
Multiplying, we get 24*2 = 48 arrangements.

Joey in Slot 5:
Since Frankie must be in Slot 6, the number of choices for Slot 6 = 1.
The number of ways to arrange the remaining 4 mobsters in Slots 1, 2, 3, and 4 = 4*3*2*1 = 24.
Multiplying, we get 1*24 = 24 arrangements.

Adding the results above, we get:
120+96+72+48+24 = 360 arrangements.
Excellent!!! i was majorly confused with Joey's placement when i tried the slot method! Thanks GG!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1