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Re: How many words (with or without meaning) can be formed using all the [#permalink]
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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!
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Re: How many words (with or without meaning) can be formed using all the [#permalink]
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Please refer to the picture for the solution.



Attachment:
IMAG0110.jpg
IMAG0110.jpg [ 2.02 MiB | Viewed 5115 times ]



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Re: How many words (with or without meaning) can be formed using all the [#permalink]
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Bunuel wrote:

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 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:
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Bookmark this post to come back to this discussion for the question links - there will be 2 on Saturday and 2 on Sunday!



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JAMBOBREE OFFICIAL SOLUTION:

We do not have a direct formula to calculate the answer , but we can say

Required answer = total number of words which can be formed – number of words when the two E”s are together

Total number of words = 6! / 2! = 360

To calculate the number of ways in which two e’s are together we will group E’s together .we will get SLFI(EE)
number of ways we can arrange SLFI(EE) = 5!

Number of words when two E’s are together = 5!

Required answer = 360 – 120
= 240
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Re: How many words (with or without meaning) can be formed using all the [#permalink]
Sorry... i dont understand why we need to divide the 6! by 2! .... Is it because we have 2 E? If we have 3 E, we will divide it by 3! ?
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Re: How many words (with or without meaning) can be formed using all the [#permalink]
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minatminat wrote:
Sorry... i dont understand why we need to divide the 6! by 2! .... Is it because we have 2 E? If we have 3 E, we will divide it by 3! ?


Yes, for SELFIEE the number of arrangements would be 7!/3! because there are 7 letters out of which 3 are identical.


THEORY:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

21. Combinatorics/Counting Methods



For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.
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Re: How many words (with or without meaning) can be formed using all the [#permalink]
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Re: How many words (with or without meaning) can be formed using all the [#permalink]
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