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 →

Combination

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 682
Joined: Fri Jan 16, 2009 2:40 amFollowed by: 1 member
Thanked: 32 times

Re: Combination

by Vemuri » Sat May 23, 2009 7:54 pm
I believe this is a permutations question, not combinations because we are dealing with arrangements.

I am guessing the answer is between D & E (E & F as stated in the answer choices), but still don't have a way to determine it. Will keep trying.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 435
Joined: Sat Sep 27, 2008 2:02 pmFollowed by: 1 member
Location: San Jose, CA
Thanked: 43 times
GMAT Score:720

by dumb.doofus » Sat May 23, 2009 10:05 pm
I think the answer should be 200. Thanks Brent for pointing out Login to see the pics..

If someone has a quicker way to solve this.. that would be awesome.. please do share..

For now here is the solution:

Let's divide this into two portions:

A] When the 3 C's are placed together

Let's denote the 3 C's as R.. So the elements are AAABBR

Permutation = 6!/(3!2!) = 60 ------------------- (1)

B] When 2 of the C's are placed together

Let's again denote the 2 C's by R

so now the elements are: AAABBCR
Now again there are two cases:
1. When R is placed as the first element or the last element. In both cases permutation will be same.

So permutation for R _ _ _ _ _ _ _ or _ _ _ _ _ _ R will be

Image

P = (5*5*4*3*2*1)/(3!2!) = 50

So total permutation for first and last position = 50 + 50 = 100 ----- (2)

2. When R is placed anywhere in the middle i.e. 2nd, 3rd, 4th, 5th or 6th position.. permutation will be same..

Image

so P = (5*4*4*3*2*1)/(3!2!) = 40

total permutation = 5 * 40 = 200 --------------------- (3)

Total permutation = 8!/(3!2!3!) - (1) + (2) + (3)
= 560 - 60 + 100 + 200
= 200
Thanks Brent for pointing out. I had missed to subtract.
Last edited by dumb.doofus on Sun May 24, 2009 9:23 am, edited 1 time in total.
One love, one blood, one life. You got to do what you should.
https://dreambigdreamhigh.blocked/
https://gmattoughies.blocked/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 260
Joined: Sun Oct 12, 2008 8:10 pm
Thanked: 4 times

by PAB2706 » Sat May 23, 2009 10:55 pm
hi i solved but got the answer 350...

:roll:

as always might have done some silly mistake
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sun May 24, 2009 7:00 am
dumb.doofus wrote:I think the answer should be 360. Login to see the pics..

If someone has a quicker way to solve this.. that would be awesome.. please do share..

For now here is the solution:

Let's divide this into two portions:

A] When the 3 C's are placed together

Let's denote the 3 C's as R.. So the elements are AAABBR

Permutation = 6!/(3!2!) = 60 ------------------- (1)

B] When 2 of the C's are placed together

Let's again denote the 2 C's by R

so now the elements are: AAABBCR
Now again there are two cases:
1. When R is placed as the first element or the last element. In both cases permutation will be same.

So permutation for R _ _ _ _ _ _ _ or _ _ _ _ _ _ R will be

Image

P = (5*5*4*3*2*1)/(3!2!) = 50

So total permutation for first and last position = 50 + 50 = 100 ----- (2)

2. When R is placed anywhere in the middle i.e. 2nd, 3rd, 4th, 5th or 6th position.. permutation will be same..

Image

so P = (5*4*4*3*2*1)/(3!2!) = 40

total permutation = 5 * 40 = 200 --------------------- (3)

Total permutation = (1) + (2) + (3)
= 60 + 100 + 200
= 360
This is a valid solution, except I believe that you have solved for the number of arrangements such that 2C's are adjacent. So, we need to subtract this value from the total number of arrangements without restrictions.
The total number of arrangements without restrictions is 8!/3!3!2! = 560
So, our final answer is 560 - 360 = 200
I'll post a slightly faster approach shortly.
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 59
Joined: Fri Feb 29, 2008 9:00 am
Thanked: 4 times

by agoyal2 » Sun May 24, 2009 7:10 am
PAB2706, I am also getting 350 :roll:

Total Arrangements = 8! / (3!*2!*3!) = 560

When 2 C's are considered 1 = 7! / (3!*2!*2!) = 210
(The above also includes when all 3 C's are together...)

So total cases = 560-210 = 350 ??
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sun May 24, 2009 7:31 am
Image
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 260
Joined: Sun Oct 12, 2008 8:10 pm
Thanked: 4 times

by PAB2706 » Sun May 24, 2009 7:32 am
hey agoyal..see one mistake that we have done is we have not concidered the possibility when all 3 C's come together.. i guess 560- 210+ when all 3C's come together shud give the answer...

i dnt know why..just getting a mental block..cant figure... :cry:
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sun May 24, 2009 8:03 am
agoyal2 wrote:PAB2706, I am also getting 350 :roll:

Total Arrangements = 8! / (3!*2!*3!) = 560

When 2 C's are considered 1 = 7! / (3!*2!*2!) = 210
(The above also includes when all 3 C's are together...)

So total cases = 560-210 = 350 ??

From your calculation, I believe that you have joined two C's and are treating it as one unit. This approach is fine, however you can't treat this entity (CC) as being identical to the remaining C. Based on your calculation of 7!/3!2!2!, it looks like you have done this.

However, the CC and the C are not identical.
For example, the arrangement AACCBBAC is different from AACBBACC

Given this, we should treat CC and C as unique to get 7!/3!2! = 420
But this figure isn't correct either since CC and C aren't entirely unique since ACCCABBA is the same as ACCCABBA

So, from our 420 we need to subrtact the number of ways in which all 3 Cs are together (60 ways).

We now have 360 (420-60) ways in which we can have two Cs adjacent, and now we can subtract this from 560 as you stated earlier.
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 487
Joined: Fri Mar 27, 2009 5:49 am
Thanked: 36 times

by dtweah » Sun May 24, 2009 8:13 am
Brent Hanneson wrote:Image
Beautiful Brent!!. I fumbled the answer choices some how and omitted the 200. I discovered error but didn't want to make the correction just to see who might still derive the correct solution from raw principles.

In case you are not familiar with the formulas Brent used u can used the following prinicples.

Since The restriction concerns the Cs, forget about them and arrange the A's and B's. How many ways can we arrange 3A's and 2'B. YOu have 5 spots. If you put the 3's in the first 3 spots, you just have 2 ways for the B's so the number of ways is 5C2. Or if you put the B's first you would have 3 ways for the A's or 5C3=5C2=10. So generally the number of ways to arrange n A's and m B's is (n+m)Cn=(n+m)Cm. So we have

A A A B B

Now we want to put the c's between these. That is the only way no two C's will be adjacent. There are actually 6 spots for any C. Count the left of first A, right of last B and all spots between. This is 3 Cs vying for 6 spots or 6C3.

10 x 20=200.

Yes Brent you are a Giant!!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 260
Joined: Sun Oct 12, 2008 8:10 pm
Thanked: 4 times

by PAB2706 » Sun May 24, 2009 8:21 am
O thats right Brent..thanks ..that is the mistake I was doing.

After some thot i arrived at this point tat the total ways for C being together will be 420 as you have shown but after that i got completely blocked..cudnt think further....

thanks a lot

cheers!!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 435
Joined: Sat Sep 27, 2008 2:02 pmFollowed by: 1 member
Location: San Jose, CA
Thanked: 43 times
GMAT Score:720

by dumb.doofus » Sun May 24, 2009 9:12 am
Brent Hanneson wrote:Image
Very nice approach Brent.. Many thanks for sharing..
One love, one blood, one life. You got to do what you should.
https://dreambigdreamhigh.blocked/
https://gmattoughies.blocked/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 59
Joined: Fri Feb 29, 2008 9:00 am
Thanked: 4 times

by agoyal2 » Sun May 24, 2009 12:43 pm
Thanks Brent!!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 319
Joined: Wed Feb 04, 2009 10:32 amFollowed by: 9 members
Location: Delhi
Thanked: 84 times

by sureshbala » Sun May 24, 2009 6:58 pm
First 3 A's and 2 B's can be arranged in 5!/(3!x2!) = 10 ways.

Now there are 6 places where 3 C's can be placed so that no two of them are together. Also, since all C's look same, arranging 3 C's in 6 places is nothing but choosing 3 places from 6 places which can be done in 6C3 = 20 ways.

Hence total number of ways = 10 x 20 = 200
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 179
Joined: Mon Dec 29, 2008 4:41 am
Thanked: 2 times

by abhinav85 » Wed May 27, 2009 10:39 am
Answer is 200.

What is th OA dtweah???
Last edited by abhinav85 on Thu May 28, 2009 5:14 am, edited 1 time in total.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply