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Math Expert V
Joined: 02 Sep 2009
Posts: 58431
How many different arrangements of A, B, C, D, and E are possible wher  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 26% (02:17) correct 74% (02:15) wrong based on 61 sessions

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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?

A. 96
B. 48
C. 32
D. 30
E. 1

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examPAL Representative P
Joined: 07 Dec 2017
Posts: 1140
Re: How many different arrangements of A, B, C, D, and E are possible wher  [#permalink]

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Bunuel wrote:
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?

A. 96
B. 48
C. 32
D. 30
E. 1

This is a hard question.
We'll calculate it by calculating all the possible arrangements, subtracting the 'bad' arrangements and correcting for double-counting.
This is similar to calculating a difficult probability by calculating 1 minus the probability and is a Precise approach.

There are 5! = 120 total options to arrange the 5 letters.
The 'bad' arrangements are those that have an A and a B adjacent or a D and an E adjacent.
To calculate this we'll first 'stick' A and B together to form 'one letter'.
Then we have 4 new letters - 'AB', C, D, E.
There are 4! = 24 ways to arrange these letters
Similarly, each of the combinations of
'BA', C, D, E
A, B, C, 'DE'
A, B, C, 'ED'
has 24 options giving a total of 24*4 = 96 'bad' options.
So far we've counted 120 - 96 = 24 options.
But - we've doublecounted! We've removed arrangements with both A,B adjacent and D,E adjacent twice!
We'll add these in similarly to what we did above:
If 'AB' and 'DE' are both together then we have 3 'letters' we can rearrange:
'AB', C, 'DE' : 3! = 6 options.
'AB', C, 'ED' : 3! = 6 options.
'BA', C, 'DE' : 3! = 6 options.
'BA', C, 'ED' : 3! = 6 options.
So we need to add back 6*4 = 24 giving a final total of 24 + 24 = 48 options.
(B) is our answer.

** It is also possible to brute force list everything but with 5 letters and all the constraints it is probably more dangerous than useful.
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Senior Manager  G
Joined: 10 Sep 2013
Posts: 307
Location: India
GMAT 1: 720 Q50 V38 GPA: 4
Re: How many different arrangements of A, B, C, D, and E are possible wher  [#permalink]

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Shouldn't it be 96??? We have and here, not or
examPAL Representative P
Joined: 07 Dec 2017
Posts: 1140
Re: How many different arrangements of A, B, C, D, and E are possible wher  [#permalink]

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darshak1 wrote:
Shouldn't it be 96??? We have and here, not or

My understanding of the original question is that (for example) 'ABDCE' is a 'bad' string because A and B are adjacent even though D and E aren't.
In this case it is a logical 'or' and not a logical 'and' even though the question says and.

I guess it depends how you read the question...
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How many different arrangements of A, B, C, D, and E are possible wher  [#permalink]

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2
We have to find the answer using negation method i.e n(with A) = n(total) - n(without A)

n(total) = 5! = 120

Here desired combo : A not adjacent to B AND D not adjacent to E
negation of this combo: ! ( A not adjacent to B AND D not adjacent to E) = A adjacent to B OR D adjacent to E

n(A adjacent to B OR adjacent to E) = n(A adjacent to B) + n(D adj to E) - n(A adj to B AND D adj to E)

n(A adjacent to B) = 4! * 2! = 48 (considering AB as single person, 4! (permutation of 4 members), 2! for (AB,BA))
n(D adjacent to E) = 4! * 2! = 48 (same logic as above)
n(A adj to B AND D adj to E) = 3! * 2! * 2! = 24 (considering AB, CD as single persons, 3!(permutation of 3 members) * 2! (for AB,BA) * 2!(for CD,DC))

n(A adjacent to B OR adjacent to E) = 48 + 48 - 24 = 72

So n(A not adjacent to B AND D not adjacent to E) = n(total) - n(A adjacent to B OR adjacent to E) = 120 - 72 = 48 => (B) How many different arrangements of A, B, C, D, and E are possible wher   [#permalink] 17 Jan 2018, 12:52
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