How many different arrangements of A, B, C, D, and E are pos : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

It is currently 23 Feb 2017, 23:45
GMAT Club Tests

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

How many different arrangements of A, B, C, D, and E are pos

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

2 KUDOS received
Director
Director
User avatar
Joined: 07 Jun 2004
Posts: 612
Location: PA
Followers: 5

Kudos [?]: 734 [2] , given: 22

How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 22 Dec 2010, 07:24
2
This post received
KUDOS
18
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

39% (02:55) correct 61% (01:48) wrong based on 345 sessions

HideShow timer Statistics

How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?

(A) 96
(B) 60
(C) 48
(D) 36
(E) 17
[Reveal] Spoiler: OA

_________________

If the Q jogged your mind do Kudos me : )

Expert Post
7 KUDOS received
Math Expert
User avatar
G
Joined: 02 Sep 2009
Posts: 37104
Followers: 7251

Kudos [?]: 96463 [7] , given: 10751

Re: Counting PS [#permalink]

Show Tags

New post 22 Dec 2010, 07:58
7
This post received
KUDOS
Expert's post
10
This post was
BOOKMARKED
rxs0005 wrote:
How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?

(A) 96
(B) 60
(C) 48
(D) 36
(E) 17


Total # of permutation of 5 distinct letters will be 5!=120;

Glue A and B together, consider it to be one unit: {AB}{C}{D}{E} --> # of permutation of these 4 units will be 4!=24, A and B within its unit also can be arranged in 2 ways : {AB} or {BA}, so total # of ways to arrange A, B, C, D, and E so that A and B to be together will be 4!*2=48;

The same for A and D: total # of ways to arrange A, B, C, D, and E so that A and D to be together will be 4!*2=48;

Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} --> # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12;

The # of arrangements when A is adjacent to neither B nor D will be total-(48+48-12)=120-84=36.

Answer: D.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
User avatar
Joined: 19 Aug 2010
Posts: 76
Followers: 3

Kudos [?]: 25 [0], given: 2

Re: Counting PS [#permalink]

Show Tags

New post 22 Dec 2010, 08:00
rxs0005,

What is the source of your questions from today?
Expert Post
1 KUDOS received
Math Expert
User avatar
G
Joined: 02 Sep 2009
Posts: 37104
Followers: 7251

Kudos [?]: 96463 [1] , given: 10751

Re: Counting PS [#permalink]

Show Tags

New post 22 Dec 2010, 08:27
1
This post received
KUDOS
Expert's post
2
This post was
BOOKMARKED
rxs0005 wrote:
How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?

(A) 96
(B) 60
(C) 48
(D) 36
(E) 17


Another way: as A must be adjacent to neither B nor D then it must be adjacent to only C or only E or both.

Adjacent to both: {CAE}{B}{D} --> # of permutation of these 3 units will be 3!, {CAE} also can be arranged in 2 ways: {CAE} or {EAC}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both C and E will be 3!*2=12;

Adjacent to only C: AC-XXX (A is the first letter and C is the second): these X-s can be arranged in 3! ways. Now, it can also be XXX-CA (A is the last letter and C is the fourth): again these X-s can be arranged in 3! ways. So total # of ways to arrange A, B, C, D, and E so that A is adjacent to only C is 3!*2=12;

The same will be when A is adjacent to only E: 3!*2=12;

Total: 12+12+12=36.

Answer: D.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Expert Post
Math Expert
User avatar
G
Joined: 02 Sep 2009
Posts: 37104
Followers: 7251

Kudos [?]: 96463 [0], given: 10751

Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 30 Jun 2013, 23:59
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

1 KUDOS received
Intern
Intern
avatar
Joined: 05 May 2013
Posts: 27
GMAT 1: 730 Q50 V39
GRE 1: 1480 Q800 V680
Followers: 0

Kudos [?]: 22 [1] , given: 5

Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 01 Jul 2013, 06:30
1
This post received
KUDOS
If the first letter is A, there are two options for the second letter (C or E), the remaining 3 letters can be arranged in 3! ways, thus there are 2*6=12 arrangements with A as the first letter. Similarly, if A were the last letter , there are 12 different arrangements. If A were in any one of the 3 intermediate positions, it wld have to be in between C and E - thus there would be 2 (betn C and E) *2 (betn B and D) =4 different arrangements for each of the 3 intermediate positions. Thus total number of diff arrangements = 12 (A is the first letter) + 12 (A is the last letter) + 4*3 (in any of the 3 intermediate posns) = 36.
Intern
Intern
avatar
Joined: 15 Apr 2014
Posts: 11
Concentration: Strategy, Marketing
GMAT 1: 700 Q49 V36
WE: Marketing (Advertising and PR)
Followers: 0

Kudos [?]: 5 [0], given: 21

Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 27 Apr 2014, 12:08
vs129 wrote:
If the first letter is A, there are two options for the second letter (C or E), the remaining 3 letters can be arranged in 3! ways, thus there are 2*6=12 arrangements with A as the first letter. Similarly, if A were the last letter , there are 12 different arrangements. If A were in any one of the 3 intermediate positions, it wld have to be in between C and E - thus there would be 2 (betn C and E) *2 (betn B and D) =4 different arrangements for each of the 3 intermediate positions. Thus total number of diff arrangements = 12 (A is the first letter) + 12 (A is the last letter) + 4*3 (in any of the 3 intermediate posns) = 36.



...thats right slot method is the way to go for solving this question ...
_________________

May everyone succeed in their endeavor. God Bless!!!

Current Student
User avatar
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 64

Kudos [?]: 604 [0], given: 355

GMAT ToolKit User
Re: Counting PS [#permalink]

Show Tags

New post 21 May 2014, 05:33
Bunuel wrote:
rxs0005 wrote:
How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?

(A) 96
(B) 60
(C) 48
(D) 36
(E) 17


Total # of permutation of 5 distinct letters will be 5!=120;

Glue A and B together, consider it to be one unit: {AB}{C}{D}{E} --> # of permutation of these 4 units will be 4!=24, A and B within its unit also can be arranged in 2 ways : {AB} or {BA}, so total # of ways to arrange A, B, C, D, and E so that A and B to be together will be 4!*2=48;

The same for A and D: total # of ways to arrange A, B, C, D, and E so that A and D to be together will be 4!*2=48;

Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} --> # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12;

The # of arrangements when A is adjacent to neither B nor D will be total-(48+48-12)=120-84=36.

Answer: D.


Looks good only thing I got wrong was that on the last step namely:

'Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} --> # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12;'

I subtracted 12 twice that is 24, because BAD and DAB are included in both the first case with AB together and the second case with AD together

Could you please explain why you only subtract once and not twice, I've made this error several times already and I can't seem to get the grip on this issue

Thanks!
Cheers
J :)
Expert Post
Math Expert
User avatar
G
Joined: 02 Sep 2009
Posts: 37104
Followers: 7251

Kudos [?]: 96463 [0], given: 10751

Re: Counting PS [#permalink]

Show Tags

New post 21 May 2014, 05:57
jlgdr wrote:
Bunuel wrote:
rxs0005 wrote:
How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?

(A) 96
(B) 60
(C) 48
(D) 36
(E) 17


Total # of permutation of 5 distinct letters will be 5!=120;

Glue A and B together, consider it to be one unit: {AB}{C}{D}{E} --> # of permutation of these 4 units will be 4!=24, A and B within its unit also can be arranged in 2 ways : {AB} or {BA}, so total # of ways to arrange A, B, C, D, and E so that A and B to be together will be 4!*2=48;

The same for A and D: total # of ways to arrange A, B, C, D, and E so that A and D to be together will be 4!*2=48;

Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} --> # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12;

The # of arrangements when A is adjacent to neither B nor D will be total-(48+48-12)=120-84=36.

Answer: D.


Looks good only thing I got wrong was that on the last step namely:

'Now, the above 48+48=96 cases will contain the arrangements when A is adjacent to both B and D, so we should subtract this cases to get rid of the double counting. The # of case when A is adjacent to both B and D will be: consider {BAD} {C}{E} --> # of permutation of these 3 units will be 3!, {BAD} also can be arranged in 2 ways: {BAD} or {DAB}, so total # of ways to arrange A, B, C, D, and E so that A is adjacent to both B and D will be 3!*2=12;'

I subtracted 12 twice that is 24, because BAD and DAB are included in both the first case with AB together and the second case with AD together

Could you please explain why you only subtract once and not twice, I've made this error several times already and I can't seem to get the grip on this issue

Thanks!
Cheers
J :)


(a) The cases for which A and B are together (48) include the cases cases when A is adjacent to both B and C: {the cases when A is adjacent only to B} + {the cases when A is adjacent to both B and C}.

(b) The cases for which A and C are together (48) include the cases cases when A is adjacent to both B and C: {the cases when A is adjacent only to C} + {the cases when A is adjacent to both B and C}.

(c) The number of cases when A is adjacent to both B and C is 12.

Now, to get the number of cases for which A is adjacent to B, or C or both = {the cases when A is adjacent only to B} + {the cases when A is adjacent only to C} + {the cases when A is adjacent to both B and C}, which is (a) + (b) - (c).

Basically the same way as when we do for overlapping sets when we subtract {both}: {total} = {group 1} + {group 2} - {both}.

Does this make sense?
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Current Student
User avatar
Joined: 06 Sep 2013
Posts: 2035
Concentration: Finance
GMAT 1: 770 Q0 V
Followers: 64

Kudos [?]: 604 [0], given: 355

GMAT ToolKit User
Re: Counting PS [#permalink]

Show Tags

New post 21 May 2014, 07:16
Oh ok, gotcha. Yeah the thing is that when in overlapping sets you only want to count the members of Set A or B, but not both then it is correct to subtract 'Both' two times.

Say like How many of the multiples of 3 and 5 are not multiples of 15?

Then you would only take the multiples of 3 and 5 and subtract 2* (Multiples of 15).

This reasoning doesn't quite apply to this question as we do in fact want to consider the scenario in which all three are seated together.
Therefore, we should use {both}: {total} = {group 1} + {group 2} - {both} as you correctly mentioned

Clear now
Thanks
Cheers!
J :)
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 13942
Followers: 590

Kudos [?]: 167 [0], given: 0

Premium Member
Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 03 Jun 2015, 02:46
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Expert Post
1 KUDOS received
SVP
SVP
User avatar
B
Joined: 08 Jul 2010
Posts: 1512
Location: India
GMAT: INSIGHT
WE: Education (Education)
Followers: 75

Kudos [?]: 1503 [1] , given: 42

How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 03 Jun 2015, 03:49
1
This post received
KUDOS
Expert's post
rxs0005 wrote:
How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?

(A) 96
(B) 60
(C) 48
(D) 36
(E) 17


ALTERNATE METHOD:

We can make cases here

Case 1: A takes position no.1 i.e. Arrangement looks like (A _ _ _ _)

In this case B and D can take any two position out of position no.3, 4, and 5
i.e. B and D can take position in 3x2 = 6 ways
remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways

i.e. total arrangement as per Case 1 = 6 x 2 = 12 ways

Case 2: A takes position no.2 i.e. Arrangement looks like (_ A _ _ _)

In this case B and D can take any two position out of position no. 4 and 5
i.e. B and D can take position in 2! = 2 ways
remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways

i.e. total arrangement as per Case 2 = 2 x 2 = 4 ways

Case 3: A takes position no.3 i.e. Arrangement looks like (_ _ A _ _)

In this case B and D can take any two position out of position no. 1 and 5
i.e. B and D can take position in 2! = 2 ways
remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways

i.e. total arrangement as per Case 3 = 2 x 2 = 4 ways

Case 4: A takes position no.4 i.e. Arrangement looks like (_ _ _ A _) This case is same as Case 2 (Just mirror of case 2) hence total ways of arrangement will remain 4 ways only

In this case B and D can take any two position out of position no. 1 and 2
i.e. B and D can take position in 2! = 2 ways
remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways

i.e. total arrangement as per Case 4 = 2 x 2 = 4 ways

Case 5: A takes position no.5 i.e. Arrangement looks like (_ _ _ _ A)This case is same as Case 2 (Just mirror of case 1) hence total ways of arrangement will remain 4 ways only

In this case B and D can take any two position out of position no.1, 2, and 3
i.e. B and D can take position in 3x2 = 6 ways
remaining two letters C and E can be arranged on remaining two places in 2! ways = 2 ways

i.e. total arrangement as per Case 5 = 6 x 2 = 12 ways

Total Ways of favorable arrangements = 12+4+4+4+12 = 36 ways

Answer: Option
[Reveal] Spoiler:
D

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com
Call us : +91-9999687183 / 9891333772
http://www.GMATinsight.com/testimonials.html



Feel free to give a Kudos if it is a useful post .

Intern
Intern
avatar
Joined: 24 Mar 2015
Posts: 10
Schools: ISB '16
Followers: 3

Kudos [?]: 1 [0], given: 10

Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 09 Sep 2015, 00:26
good question...120-48-48+(3!*2)=36
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 13942
Followers: 590

Kudos [?]: 167 [0], given: 0

Premium Member
Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 17 Sep 2016, 00:58
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Manager
Manager
avatar
S
Joined: 21 Apr 2016
Posts: 173
Followers: 0

Kudos [?]: 10 [0], given: 67

GMAT ToolKit User Reviews Badge
How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 22 Oct 2016, 06:36
A will be adjacent to B or D, in any of the permutations of ABD.

So total permutations of "CE[ABD]" is 3! * 3!

Total permutations of "ABCDE" is 5!.

Shouldn't the answer be 5! - 3! * 3!. What is it that I'm missing ?
Senior Manager
Senior Manager
User avatar
B
Status: Exam scheduled!!
Joined: 05 Sep 2016
Posts: 411
Location: United States (WI)
Concentration: Marketing, Technology
WE: Other (Law)
Followers: 3

Kudos [?]: 13 [0], given: 282

Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 17 Nov 2016, 20:11
I'm a visual person, so maybe my explanation will help those who are having difficulty seeing what's going on in this problem:
Given that we have 5 different letters to order, there are 5 places A could be situated - thus, we will devise 5 cases

(1) A __ __ __ __
(2) __ A __ __ __
(3) __ __ A __ __
(4) __ __ __ A __
(5) __ __ __ __ A

We also know that B or D cannot be adjacent to A, so we will adjust our scenarios to reflect such:

(1) A (~B/D) __ __ __
(2) (~B/D) A (~B/D) __ __
(3) __ (~B/D) A (~B/D) __
(4) __ __ (~B/D) A (~B/D)
(5) __ __ __ (~B/D) A

Now, looking at the scenarios, starting with (1), we see that position two will have to be taken by either C or E. This will give us two baseline scenarios to work off of

(a) A C __ __ __
(b) A E __ __ __

Now the remaining three slots available can be occupied by a combination of the remaining 3 letters --> 3!
Thus, for each scenario within main scenario (1) (i.e. A in first slot), we will have 12 combinations (i.e. 6 each)

This will also be the case for (5), since A will be basically flipped to the opposite side --> so we can say right now that (5) will also have 12 combinations

(2) C/E A C/E __ __ --> breaks down into two scenarios again

(a) C A E __ __ --> two combinations (with remaining letters)
(b) E A C __ __ --> two combinations (with remaining letters)

Thus (2) gives us 4 combinations

(4) will also give us 4 combinations for the same reason

(3) __ C/E A C/E __ remains --> break into scenarios once more

(a) __ C A E __ --> 2 combinations possible (with remaining letters)
(b) __ E A C __ --> 2 combinations possible (with remaining letters)

Thus, adding (1)+(2)+(3)+(4)+(5) = 36 combinations
Expert Post
1 KUDOS received
Veritas Prep GMAT Instructor
User avatar
B
Joined: 16 Oct 2010
Posts: 7185
Location: Pune, India
Followers: 2167

Kudos [?]: 14019 [1] , given: 222

Re: How many different arrangements of A, B, C, D, and E are pos [#permalink]

Show Tags

New post 18 Nov 2016, 03:33
1
This post received
KUDOS
Expert's post
rxs0005 wrote:
How many different arrangements of A, B, C, D, and E are possible where A is adjacent to neither B nor D?

(A) 96
(B) 60
(C) 48
(D) 36
(E) 17


If A can be adjacent to neither B nor D, it would be either nestled between E and C or at an extreme position next to E or next to C.

No of cases with A nestled between E and C = 3! * 2! = 12

No of cases with A at left extreme with C/E next to it = 2*3! = 12 (Choose one of C or E in 2 ways and arrange the remaining 3 letters)

No of cases with A at right extreme with C/E next to it = 12 (same as above)

Total = 12+12+12 = 36

Answer (D)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199

Veritas Prep Reviews

Re: How many different arrangements of A, B, C, D, and E are pos   [#permalink] 18 Nov 2016, 03:33
    Similar topics Author Replies Last post
Similar
Topics:
Experts publish their posts in the topic How many different arrangements of the letters A, B, C, D, and E are Bunuel 3 26 Dec 2016, 10:05
2 Experts publish their posts in the topic How many different arrangements of A, B, C, D, and E are possible wher Bunuel 3 21 Dec 2016, 03:57
1 Experts publish their posts in the topic In how many different ways can the letters B, E, N, E, F, I, C, I, A, Bunuel 1 02 Oct 2016, 04:55
6 Experts publish their posts in the topic How many different arrangements of A, B, C, D, and E are possible jp001 8 02 Jun 2015, 12:32
11 Experts publish their posts in the topic In how many ways can 6 people, A, B, C, D, E, F be seated Smita04 7 07 Feb 2012, 20:34
Display posts from previous: Sort by

How many different arrangements of A, B, C, D, and E are pos

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


cron

GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.