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Jamboree and GMAT Club Contest: How many words (with or without meanin

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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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New post 18 Nov 2015, 04:13
1
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 number of possible words with two E’s not together = Total number of possible words - Total number of possible words with two E’s together

No of letters = 6
No of repeted letters = 1
No of repitition = 2

Hence Total number of possible words = 6!/2
And Total number of possible words with two E’s together = 5!

Therefore Total number of possible words with two E’s not together = 6!/2 - 5!
=5!(6/2-1)
=5!(3-1)
=5!(2)
=120*2=240

Ans: E
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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New post 20 Nov 2015, 01:06
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There are 6 letters => the number of words including 6 letters is 6! = 2*3*4*5*6 = 720 words
Since of 6 letters, E is double =>when interchanging 2 letter E, we have the same word (e.g: E(1)xxxE(2) is the same to E(2)xxx(E1) ) => the number of distinctive words is 720/2=360 words.

Since 2 letter E cannot stand together: let's consider "EE" is one letter => the number of 5-letter words (letter EE and 4 other letters) is 5! = 2*3*4*5 = 120 words.

=> the answer is 360-120=240 words =>E is the correct answer
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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New post 20 Nov 2015, 10:34
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First we have to find the quantity of ways that we can arrange the letters:
6!/2! (because there are 2 repetitions - E)
Total = 360 ways

Now we have to find how many ways we can have the E's together:
5! (We need to think that the 2 E's are only one letter)
120 ways

Final result:
360 - 120 = 240 ways
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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New post 20 Nov 2015, 11:01
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Number of words that can be formed with EE never together in SELFIE = Total number of words - Number of words in which EE are always together.

Total number of words = 6!/2!
Number of words in which EE are always together in SELFIE = 5!
mber of words that can be formed with EE never together in SELFIE = 6!/2! - 5! = 360 -120 = 240
So answer is choice E
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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New post 20 Nov 2015, 14:50
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Answer = E
There are 6 letters with 2 repetitions. Total number of ways to arrange letters to form distinct words: 6!/2! = 360
Applying the rule, we don't want the two Es together. We calculate the number of ways to have the Es together and subtract from the total. Placing the Es together is the same as thinking of them as the same letter. The question then becomes how do you arrange 5 letters (the four non-Es and the one 2-E combination) = 5!. Therefore answer = 6!/2! - 5! = 240
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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New post 28 Nov 2015, 08:55
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:


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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: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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New post 09 Jun 2017, 10:05
\(\frac{6!}{2!} - 5!\) = 240
Answer E
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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New post 27 Feb 2018, 00:46
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: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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New post 27 Feb 2018, 01:00
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: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin   [#permalink] 10 May 2019, 04:18

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