GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Oct 2019, 03:30

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

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

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

Hide Tags

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

Show Tags

New post 25 Dec 2017, 01:49
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

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

HideShow timer Statistics

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

Show Tags

New post 25 Dec 2017, 03:10
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.
_________________
Senior Manager
Senior Manager
User avatar
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]

Show Tags

New post 25 Dec 2017, 05:53
Shouldn't it be 96??? We have and here, not or
examPAL Representative
User avatar
P
Joined: 07 Dec 2017
Posts: 1140
Re: How many different arrangements of A, B, C, D, and E are possible wher  [#permalink]

Show Tags

New post 25 Dec 2017, 07:29
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...
_________________
Senior Manager
Senior Manager
avatar
P
Joined: 02 Apr 2014
Posts: 468
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34
GPA: 3.5
How many different arrangements of A, B, C, D, and E are possible wher  [#permalink]

Show Tags

New post 17 Jan 2018, 12:52
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)
GMAT Club Bot
How many different arrangements of A, B, C, D, and E are possible wher   [#permalink] 17 Jan 2018, 12:52
Display posts from previous: Sort by

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

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





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne