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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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;

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

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

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

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}.

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

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

How many different arrangements of A, B, C, D, and E are pos [#permalink]
03 Jun 2015, 03:49

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

Contact for One-on-One LIVE Online (SKYPE Based) Quant/Verbal Demo ______________________________________________________ Please press the if you appreciate this post !!

gmatclubot

How many different arrangements of A, B, C, D, and E are pos
[#permalink]
03 Jun 2015, 03:49

The Stanford interview is an alumni-run interview. You give Stanford your current address and they reach out to alumni in your area to find one that can interview you...

Originally, I was supposed to have an in-person interview for Yale in New Haven, CT. However, as I mentioned in my last post about how to prepare for b-school interviews...

Interested in applying for an MBA? In the fourth and final part of our live QA series with guest expert Chioma Isiadinso, co-founder of consultancy Expartus and former admissions...