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

Redeem

combinatronics problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 496
Joined: Tue Jun 07, 2011 5:34 am
Thanked: 38 times
Followed by:1 members

combinatronics problem

by sl750 » Thu Jul 21, 2011 12:55 pm
In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the
letter C must be to the right of the letter D?

1680
2160
2520
3240
3360

I didn't follow the explanation given for this problem. The solution is A
Top
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 613
Joined: Thu Mar 22, 2007 6:17 am
Location: madrid
Thanked: 171 times
Followed by:64 members
GMAT Score:790

by kevincanspain » Thu Jul 21, 2011 1:17 pm
First, let's ignore the restriction, as for every distinct way there is of arranging the letters so that C is to the right of D, there is exactly one way of arranging the letters so that C is to the left of D.

The number of ways of arranging n elements in a row if k are identical and the other n - k are distinct is n!/k!

For example, the letters of Madrid can be arranged in 6!/2! ways, the letters of Malaga in 6!/3! ways, and those of Zaragoza in 8!/2!3! ways.

2 A's 3 B's 1 C 1 D and 1 E can be arranged in 8!/2!3! ways = 8 x 7 x 6 x 5 x 4/2 = 56 x 60

In exactly half of these linear arrangements, 56 x 30 , the C is to the right of D
Kevin Armstrong
GMAT Instructor
Gmatclasses
Madrid
Top
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pm
Thanked: 114 times
Followed by:11 members

by knight247 » Thu Jul 21, 2011 8:40 pm
Hey Kevin,
I've come across many similar problems wherein among the letters A,A,B,B,B,C,D,E ......D,C can be replaced by X...Making the letters A,A,B,B,B,X,E and then arranged in 7!/2!3!.But I'm getting an answer of 480 so I'm missing a 4 somewhere...Also, I've obviously not arranged D,C in 2 ways coz they have to stay in that position.... I'm not able to crack this problem using that method of solving. Just hoping u show me how to do it in the above way..Thanks
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 496
Joined: Tue Jun 07, 2011 5:34 am
Thanked: 38 times
Followed by:1 members

by sl750 » Thu Jul 21, 2011 9:42 pm
Thanks for the reply Kevin.

I didn't follow the last bit of the explanation. i.e, how in exactly half of the arrangements will C be to the right of D

Thanks
Top
Senior | Next Rank: 100 Posts
Posts: 54
Joined: Wed Jul 20, 2011 12:36 pm
Thanked: 2 times

by sumasajja » Fri Jul 22, 2011 12:41 am
i analysed the topic n found that.... - - - - - - - - suppose we try 2 make 2 letters together then there will be 7 ways to do that for example
cd------
-cd-----
--cd----
---cd---
----cd--
-----cd-
------cd
isn't my way of approach to the problem correct??? or is it wrong!!!...please answer
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pm
Location: Milpitas, CA
Thanked: 1854 times
Followed by:523 members
GMAT Score:770

by Anurag@Gurome » Fri Jul 22, 2011 2:28 am
sl750 wrote:In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the
letter C must be to the right of the letter D?
The problem doesn't say that C must be positioned on the immediate right of D. C can be placed anywhere on the left of D. For example, AABCEBD or CAABBBED etc. Hence the approach of binding the letters together doesn't apply here.

The letters can be arranged among themselves in 8!/(3!*2!) = 3360 ways.
Now, in each of these arrangements C is either on the left of D or on the right. Hence, in exactly half of the arrangements C will be on the right of D.

Thus, number of arrangements such that C is on the right of D = 3360/2 = 1680

The correct answer is A.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 496
Joined: Tue Jun 07, 2011 5:34 am
Thanked: 38 times
Followed by:1 members

by sl750 » Fri Jul 22, 2011 3:39 am
Ok. I see it now. Thanks Anurag!!
Top
Senior | Next Rank: 100 Posts
Posts: 54
Joined: Wed Jul 20, 2011 12:36 pm
Thanked: 2 times

by sumasajja » Fri Jul 22, 2011 9:44 am
Anurag@Gurome wrote:
sl750 wrote:In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the
letter C must be to the right of the letter D?
The problem doesn't say that C must be positioned on the immediate right of D. C can be placed anywhere on the left of D. For example, AABCEBD or CAABBBED etc. Hence the approach of binding the letters together doesn't apply here.

The letters can be arranged among themselves in 8!/(3!*2!) = 3360 ways.
Now, in each of these arrangements C is either on the left of D or on the right. Hence, in exactly half of the arrangements C will be on the right of D.

Thus, number of arrangements such that C is on the right of D = 3360/2 = 1680

The correct answer is A.

Thanks MR.Anurag ....n i really appreciate ur help .if the question asked is to find number of arrangements such that d is on the right of c........would the answer be same as the above answer.....i need a little clarification on that
Top
Post new topic Post Reply
• Page 1 of 1