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

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How many different arrangements of A, B, C, D, and E are pos [#permalink] New post 22 Dec 2010, 07:24
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A
B
C
D
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Question Stats:

36% (02:59) correct 63% (01:40) wrong based on 130 sessions
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

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Re: Counting PS [#permalink] New post 22 Dec 2010, 07:58
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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.
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Re: Counting PS [#permalink] New post 22 Dec 2010, 08:00
rxs0005,

What is the source of your questions from today?
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Re: Counting PS [#permalink] New post 22 Dec 2010, 08:27
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


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.
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Re: How many different arrangements of A, B, C, D, and E are pos [#permalink] New post 30 Jun 2013, 23:59
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: How many different arrangements of A, B, C, D, and E are pos [#permalink] New post 01 Jul 2013, 06:30
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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.
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Re: How many different arrangements of A, B, C, D, and E are pos [#permalink] 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 ...
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Re: Counting PS [#permalink] 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 :)
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Re: Counting PS [#permalink] New post 21 May 2014, 05:57
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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?
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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.


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Re: Counting PS [#permalink] 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 :)
Re: Counting PS   [#permalink] 21 May 2014, 07:16
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