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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Difficult Math Question #31 - Combinations

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members
Location: Arabian Sea
Thanked: 125 times

by ajith » Sun Oct 22, 2006 9:42 pm
Long Method
As there are only 5 participants we can split this into four cases

1. 3 People between them

There are 3! ways for this to happen
6 ways

2. 2 People between them

We can select the man who is gonna stand out in 3c1 ways and arrange him either infront of the group or after the group. Again you can arrange the people between them in 2 ways

so 2*3c1*2= 12 ways

3. 1 between them

We can select the one who is gonna be between them in 3c1 ways
then we can arrange these people in 3!( two other men and the group)

So the total no: of ways =18 (3!*3)

4. None between them

we can arrange the people in 4! ways ( 3 people and a group)

that is 24 ways


So the total no: of ways.. 24+18+12+6=60

Simpler Method

As there are 5 people there are 5! ways to arrange them
ie 120 ways to arrange them ...

So, in half of them one guy has to be ahead of the other and vice versa

that gives us 60 as answer
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 354
Joined: Tue Jun 27, 2006 9:20 pmFollowed by: 5 members
Thanked: 11 times

OA

by 800guy » Mon Oct 23, 2006 2:26 pm
OA:

approach: first fix Jen, and then fix Bob. Then fix the remaining three.

Case 1: When Jen is in the first place.
=> Bob can be in any of the other four places. => 4.
The remaining 3 can arrange themselves in the remaining 3 places in 3! Ways.
Hence total ways = 4*3!

Case 2: When Jen is in the second place.
=> Bob can be in any of three places => 3.
The remaining can arrange themselves in 3 places in 3! Ways.

Continuing the approach.
Answer = 4*3! +3*3! +2*3! +1*3! = 10*3! = 60 ways.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 57
Joined: Wed Jul 12, 2006 3:13 pmFollowed by: 4 members

by limits660 » Wed Oct 25, 2006 4:36 am
I have a question about this one

here are what i believe to be the possibilities:

1 2 3 4 5
J B X X X
J X B X X
J X X B X
J X X X B
X J B X X
X J X B X
X J X X B
X X J B X
X X J X B
X X X J B

Therefore 10 ways that Jan finished in front of Bob

How can there be 60? Im missing something fundamental, right?
-
Jeff Sacco
www.jeffsacco.ca
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1275
Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members
Location: Arabian Sea
Thanked: 125 times

by ajith » Wed Oct 25, 2006 4:55 am
limits660 wrote:I have a question about this one

here are what i believe to be the possibilities:

1 2 3 4 5
J B X X X
J X B X X
J X X B X
J X X X B
X J B X X
X J X B X
X J X X B
X X J B X
X X J X B
X X X J B

Therefore 10 ways that Jan finished in front of Bob

How can there be 60? Im missing something fundamental, right?

The problem here is X's you used are not identical, They can be arranged in 6 ways each.

For example Let us assume that 5 members are X,Y,Z, J ,B

then we will take your first arrangement,

J B X Y Z

this is not the only one

J B X Z Y
J B Z X Y
J B Z Y X
J B Y X Z
J B Y Z X

are also there,ie you missed 5 of those...

For each of your arrangement there are 5 more are there ..

That gives you 10*5 =50 more arrangements...

50+10 is 60 ....


Hope that helps!
:)
Last edited by ajith on Wed Oct 25, 2006 7:01 am, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 57
Joined: Wed Jul 12, 2006 3:13 pmFollowed by: 4 members

by limits660 » Wed Oct 25, 2006 5:58 am
Ahhh yes

good point

now I see my flaw. Thank you very much for the explaination
-
Jeff Sacco
www.jeffsacco.ca
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 » Mon Mar 30, 2009 6:37 pm
5!/2 = 60
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1