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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
62% (01:29) correct
38%
(01:44)
wrong
based on 423
sessions
History
Date
Time
Result
Not Attempted Yet
How many words (with or without meaning) can be formed using all the letters of the word “SELFIE” so that the two E’s are not together?
(A) 660
(B) 600
(C) 500
(D) 300
(E) 240
QUESTION #10:
For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend.
To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.
PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:
-- GMAT Online Comprehensive (If the student wants an online GMAT preparation course)
-- GMAT Classroom Program (Only if he/she has a Jamboree center nearby and is willing to join the classroom program)
Bookmark this post to come back to this discussion for the question links - there will be 2 on Saturday and 2 on Sunday!
There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.
All announcements and winnings are final and no whining
GMAT Club reserves the rights to modify the terms of this offer at any time.
JAMBOBREE OFFICIAL SOLUTION
(A) 660
(B) 600
(C) 500
(D) 300
(E) 240
Jamboree and GMAT Club Contest Starts
QUESTION #10:
For the following two weekends we'll be publishing 4 FRESH math questions and 4 FRESH verbal questions per weekend.
To participate, you will have to reply with your best answer/solution to the new questions that will be posted on Saturday and Sunday at 9 AM Pacific.
Then a week later, respective forum moderators will be selecting 2 winners who provided most correct answers to the questions, along with best solutions. Those winners will get 6-months access to GMAT Club Tests.
PLUS! Based on the answers and solutions for all the questions published during the project ONE user will be awarded with ONE Grand prize. He/She can opt for one of the following as a Grand Prize. It will be a choice for the winner:
-- GMAT Online Comprehensive (If the student wants an online GMAT preparation course)
-- GMAT Classroom Program (Only if he/she has a Jamboree center nearby and is willing to join the classroom program)
Bookmark this post to come back to this discussion for the question links - there will be 2 on Saturday and 2 on Sunday!
There is only one Grand prize and student can choose out of the above mentioned too options as per the conditions mentioned in blue font.
All announcements and winnings are final and no whining
GMAT Club reserves the rights to modify the terms of this offer at any time. JAMBOBREE OFFICIAL SOLUTION
14 Nov 2015, 09:52
Kudos
Bookmarks
E
= Total number of words possible - Total no of words where E is together
= (6! /2) - 5!
= 240
= Total number of words possible - Total no of words where E is together
= (6! /2) - 5!
= 240
14 Nov 2015, 13:12
Kudos
Bookmarks
The question is asking the total number of arrangements possible with the letters of the word “SELFIE” where two E’s are not together.
Arrangements when two E’s are not together = Total arrangements - Arrangements when two E’s are together
In total there are 6 letters but two are identical. so we can arrange in 6! ways. but we divide for those objects that are identical. so divide by 2!. Hence,
Total arrangements = 6!/2!
Now two E's are coupled together. Consider this couple (EE) as one letter. apart from this there are 4 more letters. so we can arrange these 5 different objects in 5! ways.
Two E's can arrange themselves in 2! ways, but we divide for those objects that are identical. so divide by 2!. so arrangement for E's would be 2!/2!.
Hence, Arrangements when two E’s are together = 5! * (2!/2!)
Arrangements when two E’s are not together = 6!/2! - 5! = 5! * ( 6/2 -1 ) = 120 * 2 = 240.
Option E is correct!
Arrangements when two E’s are not together = Total arrangements - Arrangements when two E’s are together
In total there are 6 letters but two are identical. so we can arrange in 6! ways. but we divide for those objects that are identical. so divide by 2!. Hence,
Total arrangements = 6!/2!
Now two E's are coupled together. Consider this couple (EE) as one letter. apart from this there are 4 more letters. so we can arrange these 5 different objects in 5! ways.
Two E's can arrange themselves in 2! ways, but we divide for those objects that are identical. so divide by 2!. so arrangement for E's would be 2!/2!.
Hence, Arrangements when two E’s are together = 5! * (2!/2!)
Arrangements when two E’s are not together = 6!/2! - 5! = 5! * ( 6/2 -1 ) = 120 * 2 = 240.
Option E is correct!
GMAT Club reserves the rights to modify the terms of this offer at any time. 
