December 15, 2018 December 15, 2018 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT. December 15, 2018 December 15, 2018 10:00 PM PST 11:00 PM PST Get the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299)
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51218

Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 08:45
Question Stats:
54% (01:26) correct 46% (01:43) wrong based on 241 sessions
HideShow timer Statistics
Jamboree and GMAT Club Contest Starts QUESTION #10: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 Check conditions below: 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 6months 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. NOTE: Test Prep Experts and Tutors are asked not to participate. We would like to have the members maximize their learning and problem solving process.
Thank you! JAMBOBREE OFFICIAL SOLUTION
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 01 Nov 2015
Posts: 10
Concentration: General Management

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 08:52
E = Total number of words possible  Total no of words where E is together = (6! /2)  5! = 240



Intern
Joined: 04 Jan 2015
Posts: 3

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 11:25
240 : (6!/2!)  (5!)



Intern
Status: You think you cannot?? Rethink & repeat until its done  Abdul Rauf Liaqat
Joined: 09 Nov 2015
Posts: 20
Location: Pakistan
Concentration: Entrepreneurship, Strategy
GPA: 3.03
WE: Operations (Telecommunications)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 11:35
In general, repetitions are taken care of by dividing the permutation by the factorial of the number of objects that are identical. in this case, if you think logically, A,B,C,D options are not divisible by 6!. 6 is total number of letters in the word SELFIE. Only option E is divisible by 6!. So correct answer is E that is 240
_________________
+1 Kudos will be appreciated if you find this post helpful Take up one idea. Make that one idea your life  think of it, dream of it, live on that idea. Let the brain, muscles, nerves, every part of your body, be full of that idea, and just leave every other idea alone. This is the way to success  Swami Vivekananda



Retired Moderator
Status: On a mountain of skulls, in the castle of pain, I sit on a throne of blood.
Joined: 30 Jul 2013
Posts: 327

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 11:41
Total ways in which "SELFIE" can be written:
6!/2!=360
When the Es are always together in "SELFIE", we have
5!=120=120
Total arrangements in which the 2 Es are not together=360120=240
Answer: E



Board of Directors
Joined: 17 Jul 2014
Posts: 2618
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 11:42
I picked E, and this is why:
Since order doesn't matter, we need to use combinatorics, not FCP. since we have 2 letters of E, we thus have overall 6!/2! options. wich is equal to 5! * 3 = 360. Note that this is the total number of combinations, including cases when 2 E's are close. Thus, we can eliminate A) 660, B) 600, and C) 500
Well, if we don't know how to solve further, at least we have a 50% chance of getting to the right answer, and make a smart guess.
Further, let's make those 2 E's as a single number, and see how many options we have when those 2 bad boys are together: we have 5! or 120 options.
Now from the total options, let's subtract the options when E's are together: 360120 = 240.
As we can see, we can eliminate D, and pick E, which is exactly 240.



Intern
Joined: 24 Oct 2013
Posts: 29
GMAT 1: 570 Q47 V23 GMAT 2: 620 Q49 V24 GMAT 3: 630 Q47 V29
WE: Operations (Energy and Utilities)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 11:45
QUESTION #10:
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
Total nos of unique words that can be arranged with 'SELFIE' is 6!/2! = 360. Out of that if we take 2 E's as one E and arrange, then it could be done in 5! = 120 ways. Therefore, total ways in which these 2 E's are not together will be 360  120 = 240 ways. Answer E



Moderator
Joined: 21 Jun 2014
Posts: 1096
Location: India
Concentration: General Management, Technology
GPA: 2.49
WE: Information Technology (Computer Software)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 12:12
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!
_________________
 Target  720740 Project PS Butler  https://gmatclub.com/forum/projectpsbutlerpracticeeveryday280904.html http://gmatclub.com/forum/informationonnewgmatesrreportbeta221111.html http://gmatclub.com/forum/listofoneyearfulltimembaprograms222103.html



Intern
Joined: 03 Nov 2015
Posts: 2

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 13:02
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  1) Let's ignore condition regarding the two E's. We have: 6!=720 words 2) Two E's cut our words by half: 720/2=360 words 3) Number of words so that E's are NOT together = 360  Number of words so that E's are together To find the Number of words with adjacent E's we should treat E's as a single letter, so we get 5!, or 120 words with adjacent E's. Number of words so that E's are NOT together = 360  120 = 240 Answer: E



Intern
Joined: 21 Jul 2015
Posts: 34

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 18:24
There are 6 letters in the word "SELFIE", out of which letter E is twice Total possible combinations with word "SELFIE" = 6!/2! = 360 words (divide by 2! since letter E repeats twice)
Total possible combinations with "EE" together in word "SELFIE" Consider "EE" as single unit: S L F I {EE} = 5 letters = 5! = 120 words
Total possible combinations with two E’s are not together = 360  120 = 240 ways Answer (E)



Verbal Forum Moderator
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2127
Location: India
Concentration: General Management, Strategy
GPA: 3.2
WE: Information Technology (Consulting)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 18:49
Total number of letters that can be formed using letters of word SELFIE = 6!/2! = 720/2= 360 (We are dividing by 2 because of 2 identical letters E) Let's consider the cases when the 2 E's are together , 2 E's can be clubbed together as One . So we have 5 letters number of words when 2 E's are together = 5! = 120 Total number of words without 2 E's together = 360120 = 240 Answer E
_________________
When everything seems to be going against you, remember that the airplane takes off against the wind, not with it.  Henry Ford The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long +1 Kudos if you find this post helpful



Math Expert
Joined: 02 Aug 2009
Posts: 7107

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
14 Nov 2015, 21:28
total ways=6!/2.. ways where both E's are together=5! ans=360120=240... E
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Intern
Joined: 12 May 2013
Posts: 2

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
15 Nov 2015, 06:42
The answer Is 240.
Number of ways for arranging "SELFIE"=6!/2!=360
number of ways where both EE are together is 5!(EE SLFI,taking EE as one)=120 required number 360120=240



Manager
Joined: 02 Nov 2014
Posts: 192
GMAT Date: 08042015

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
15 Nov 2015, 07:41
There are total 6 letters: S L F I E E with 2 E repeats. So, the total number of words possible = 6! / 2! = 360. ....(i) Now, the no. of words with EE next to each other = no. of words considering the 2E's together as a single entity = 5! = 120.......(ii) So, the no. of words with 2E's NOT TOGETHER = i  ii = 360  120 = 240.
Choice E. 240.



Manager
Status: Just redeemed Kudos for GMAT Club Test !!
Joined: 14 Sep 2013
Posts: 94
Location: Bangladesh
GPA: 3.56
WE: Analyst (Commercial Banking)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
15 Nov 2015, 08:52
in 'SELFIE', there are 6 letters with repetition of 'E' twice. Thus 'SELFIE' can be arranged in \(\frac{6!}{2}\) = \(\frac{720}{2}\) = 360 ways. Again, the number of ways the 2 'E' can be together is 5! =120. So number of words can be formed using all “SELFIE” so that the two E’s are not together = 360120 = 240. Answer: 240 (E)
_________________
______________ KUDOS please, if you like the post or if it helps "Giving kudos" is a decent way to say "Thanks"
Master with structure  Numerical comparison [source: economist.com] https://gmatclub.com/forum/masterwithstructurenumericalcomparison233657.html#p1801987



SC Moderator
Joined: 13 Apr 2015
Posts: 1687
Location: India
Concentration: Strategy, General Management
GPA: 4
WE: Analyst (Retail)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
15 Nov 2015, 19:34
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?
Total number of words that can be formed = 6!/2! = 720/2 = 360
Number of words that can be formed with 2 E's together = 5! = 120
Number of words that can be formed so that 2 E's are not together = Total  (2 E's together) = 360  120 = 240
Answer: E



Intern
Joined: 30 Aug 2015
Posts: 3
Concentration: Marketing, Other
GPA: 3.5
WE: Analyst (Consulting)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
15 Nov 2015, 22:44
Alternate Method:
SELFIE in which SLFI can be arranged in 4! ways Again the EE can be arranged in 2! ways But due to repetition of EE 4!*2!/2! = 4! The condition given is No Two EE Together so 5C2
The Total solution comes up as 4!*5c2 = 240 .
Answer is E .



Intern
Joined: 20 Aug 2015
Posts: 4

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
16 Nov 2015, 06:31
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: In how many ways can you arrange the letters 'S', 'L', 'F', 'I', 'E', 'E' so that the two E's are not put together.
Solution: There are 6 letters in total which can be arranged in 6 x 5 x 4 x 3 x 2 x 1 = 6! ways. But, this will have all the combinations with and without the two E's together, e.g. FEELIS, FELESI, etc.
Now, if we assume the two E's to be 1 block (for the combinations of letters which have the two E's together), we have 5 letters ('S', 'L', 'F', 'I', 'EE') that can be arranged in 5 x 4 x 3 x 2 x 1 = 5! ways.
Therefore, total ways the 6 letters can be arranged so that the two E's are not together = total ways to arrange the 6 letters  total ways to arrange the 6 letters with the two E's together = 6!  5! = 6x5!  5! = 5!(61) = 5! x 5 = 120 x 5 = 600
Answer: (B) 600



Intern
Joined: 02 Oct 2012
Posts: 15
Location: India
Concentration: Operations, General Management
GPA: 3.6
WE: Operations (Energy and Utilities)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
16 Nov 2015, 09:21
Please refer to the picture for the solution. Attachment:
IMAG0110.jpg [ 2.02 MiB  Viewed 2443 times ]
Saunak Dey



Current Student
Joined: 03 May 2014
Posts: 63
Concentration: Operations, Marketing
GMAT 1: 680 Q48 V34 GMAT 2: 700 Q49 V35
GPA: 3.6
WE: Engineering (Energy and Utilities)

Re: Jamboree and GMAT Club Contest: How many words (with or without meanin
[#permalink]
Show Tags
16 Nov 2015, 13:38
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?
AnsE Total Number of words which can be formed with given letters 6!/2! = 360 We divided by 2! as to take out the words those are repeated because of Two Es Now Number of words so that two Es are not together= Total Number of words with given letters minus Number of words so that two Es are together 1 Number of words so that two Es are together= EE SLFI Taking EE as one letter and then finding words=5!=120 So from equation 1. we can have 360120=240




Re: Jamboree and GMAT Club Contest: How many words (with or without meanin &nbs
[#permalink]
16 Nov 2015, 13:38



Go to page
1 2
Next
[ 29 posts ]



