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

Correct if I am wrong, I am new here on the GMAT forum

So you have A,B,C,D,E
five possible ways to arrange it = 5! = 120

AB cannot be adjacents neither DE:
(AB),C,D,E = 4! = 24 x 2(because it can be (AB) or (BA) ) = 48
A,B,C,(DE) = 4! = 24 x 2 = 48

To get rid of the double counting:
(AB),C,(DE) = 3! x 4 = 24
(Four possible ways to arrange it)

120 -48-48+24 = 48 possible ways

its more easy if you consider the case where A is adjacent to B and D is Adjacent to E and then subtracting it from total arrangements
total arrangements = 5! = 120
Question ask you for the cases where A is adjacent to B and D is to E...we have to subtract the cases where A is adjacent to B But D is not adjacent to E and vice versa
A adjacent to B only case= (AD) take this AD as one
(AB) C D E = 4! = 24
Now AB could be BA so we multiply 24 with 2= 24*2 = 48--->case one

similarly cases where D is adjacent to E but A is not Adjacent to B (same process as above)
we will get 48---case two

now their is one case that we added twice
AB C DE = 3! = 6

AB could be BA and DE could be ED so = 2*2 = 4
6*4= 24---->case three

Total - Case one - Case two + case three
120 -48 -48 + 24 = 48 possible ways
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Similar question to practice: how-many-different-arrangements-of-a-b-c-d-and-e-are-pos-106764.html
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ALTERNATE METHOD:

We could make several cases to arrange them

Case 1: A takes first position of the 5 positions i.e. Arrangement looks like (A _ _ _ _)

The possibilities here as per favourable cases condition are
A D B E C
A D B C E
A E B D C
A E B C D
A D C B E
A E C B D
A C D B E
A C E B D
A D C E B
A E C D B
10 Cases

Case 2: A takes Second position of the 5 positions i.e. Arrangement looks like (_ A _ _ _)

The possibilities here as per favourable cases condition are
D A E B C
E A D B C
D A C B E
E A C B D
C A D B E
C A E B D

D A E C B
E A D C B
D A C E B
E A C D B
10 Cases

Case 3: A takes Third position of the 5 positions i.e. Arrangement looks like (_ _ A _ _)

The possibilities here as per favourable cases condition are
D C A E B
E C A D B
C D A E B
C E A D B
and 4 more cases which will be Mirror image of the above mentioned 4 cases i.e.
B E A C D
B D A C E
B E A D C
B D A E C
8 Cases

Case 4: A takes Forth position of the 5 positions i.e. Arrangement looks like (_ _ _ A _)[This case is mirror image of Case 2 hence same no. of arrangements will be obtained which will all be mirror image of arrangements of Case 2]
10 Cases

Case 5: A takes Fifth position of the 5 positions i.e. Arrangement looks like (_ _ _ _ A)[This case is mirror image of Case 1 hence same no. of arrangements will be obtained which will all be mirror image of arrangements of Case 1]
10 Cases

Total Arrangements = 10 + 10 + 8 + 10 + 10 = 48 Arrangements

Answer: Option

Can someone please help, as Iam unable to understand why are we considering an 'OR' condition here when the question says
where A is not adjacent to B and D is not adjacent to E?

iam guessing the solution should be 5! - the number of cases when AB and DE are together
i.e 120 - (3!*2!*2!) = 96
Please correct me

Hi Kasturi,

The question has mentioned that Neither should A be next to B nor should D be next to E, whereas your solution find the cases where both A and B are together as well as D and E are together but your solution does NOT exclude

1) the cases where A and B are together but D and E are NOT together
2) the cases where D and E are together but A and B are NOT together

hence your solution is INCORRECT

I hope it clears your doubt!!!

I dont get it , stem asks How many different arrangements of A, B, C, D, and E are possible where A is not adjacent to B and D is not adjacent to E?

and not OR, , having said that in my opinion sol should be 5! - 3!*4 = 96 option A

Bunuel Can you please explain why we are considering an OR scenario. We should find out the no. of cases where A and B are together as well as D and E are together.

Hence AB C DE = 3! *2*2 = 24

This should now be subtracted from 120 and the answer should be 96.

Can you please explain what am i missing here?
Thanks.!

I have the exact same question.
Can Bunuel/ VeritasPrepKarishma / mikemcgarry / Abhishek009 /
chetan2u or any other expert please confirm why we are using the Or condition when the question clearly states:

How many different arrangements of A, B, C, D, and E are possible where A is not adjacent to B and D is not adjacent to E?

If the question was :

How many different arrangements of A, B, C, D, and E are possible where neither is A adjacent to B nor is D adjacent to E, then the OA does make sense.

Could you please confirm if my understanding is right?
A reply will be much appreciated.

We need two cases in which X doesn't happen and neither does Y happen.
X - A is not adjacent to B
Y - D is not adjacent to E
So any case that has X but not Y should be included. Similarly, any case that has Y but not X should also be included.
Hence Or is the correct way.

Had the question been "X and Y do not happen simultaneously" then we would have used "and".

Hi Karishma
Appreciate your quick response!
Unfortunately, I am not able to follow what you are trying to say. Is there any other way you could help me understand this better?

Dear darn,

I'm happy to respond.

My friend, I don't know how much formal logic you have studied. Students sometimes naively treat the word "not" as the equivalent of a negative sign and treat the word "and" as the equivalent of multiplication, and then try to apply the Distributive Law. The problem, though, is that the word "not" does not distribute they way that numbers would.

(not A) and (not B) \(\neq\) not (A and B)

Let's think about a real world example. Let's say I have a company and I am hiring. Let's say A = no college education, so I can't hire somebody with condition A. Let's say B = constantly covered in insects, so I can't hire somebody with B (it's unclear whether this would be construed as discrimination under current labor laws).

Suppose I post a sign #1: We hire only people who are not A and not B.
Person #1 has no college education but no insects (A and not B)
Person #2 has a college education but is constantly covered in insect (B and not A)
Person #3 has no college education and is constantly covered in insect (not A and not B)
All three people are automatically excluded by that sign, because that sign specifies two conditions that must be met simultaneous--the "not A" condition and the "not B" condition. We ask each person two questions: are you A? are you B? Both have to be "no" answers to be included. Each of these three people would say "yes" to at least one of the questions, so all three would be excluded.

Instead, suppose I post a sign #2: We hire only people who are not A and B.
Here, we are excluding simply one condition, the condition of having both A & B simultaneously. We ask each person just one question: are you both A and B at the same time? Persons #1 and #2 can truthfully say no, because each of them has only one of the conditions, and only person #3 has to say yes. Thus, this sign still excludes poor person #3, but now person #1 and person #2 would be included, because neither one of them has the combined condition of A & B simultaneously. This has a different effect than sign #1 had!

In fact, and this is the part that can be really confusing for people who haven't studied formal logic,
(not A) and (not B) = not (A or B)
Similarly,
(not A) or (not B) = not (A and B)

Think about a third possible sign, sign #3, saying: We hire only people who are not A or B.
Here, again, we are excluding just one condition, but now it's the condition of A or B. Again, we ask each person just one question: are you either A or B? All three people would have to respond yes to this question, so all three would be excluded.

The same people are excluded by sign #1 and sign #3, so those two signs have the same meaning, but sign #2 has a different meaning.

The folks who solved this math problem correctly used these logical principles correctly.

Does all this make sense?
Mike
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Without any restrictions, there are 5! = 120 arrangements. Of these arrangements, 2! x 4! = 2 x 24 = 48 arrangements have A and B next to each other and another 48 arrangements have C and D next to each number. However, there are 2! x 2! x 3! = 2 x 2 x 6 = 24 arrangements that have A and B next to each other AND C and D next to each number. Therefore, the number of arrangements where A and B next to each other OR C and D next to each number is 48 + 48 - 24 = 72.

Since the number of arrangements where A is not adjacent to B and D is not adjacent to E is equal to the total number of arrangements (without any restrictions) minus the number of arrangements where A and B are next to each other OR C and D are next to each number, that number is 120 - 72 = 48.

Answer: B
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