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

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

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14 Nov 2015, 08:45
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10
00:00

Difficulty:

55% (hard)

Question Stats:

54% (01:26) correct 46% (01:43) wrong based on 241 sessions

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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

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

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14 Nov 2015, 08:52
3
E
= Total number of words possible - Total no of words where E is together
= (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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14 Nov 2015, 11:25
1
240 : (6!/2!) - (5!)
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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14 Nov 2015, 11:35
1
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
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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14 Nov 2015, 11:41
2
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=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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14 Nov 2015, 11:42
1
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:
360-120 = 240.

As we can see, we can eliminate D, and pick E, which is exactly 240.
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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14 Nov 2015, 11:45
1
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.
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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14 Nov 2015, 12:12
1
1
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: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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14 Nov 2015, 13:02
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
---------
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
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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14 Nov 2015, 18:24
1
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
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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14 Nov 2015, 18:49
1
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 = 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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14 Nov 2015, 21:28
1
total ways=6!/2..
ways where both E's are together=5!
ans=360-120=240...
E
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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15 Nov 2015, 06:42
1

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 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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15 Nov 2015, 07:41
1
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.
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Re: Jamboree and GMAT Club Contest: How many words (with or without meanin  [#permalink]

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15 Nov 2015, 08:52
1
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 = 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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15 Nov 2015, 19:34
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?

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

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

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15 Nov 2015, 22:44
1
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 .

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

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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', 'E-E') 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!(6-1)
= 5! x 5
= 120 x 5
= 600

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

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16 Nov 2015, 09:21
1
Please refer to the picture for the solution.

Attachment:

IMAG0110.jpg [ 2.02 MiB | Viewed 2443 times ]

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

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16 Nov 2015, 13:38
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?

Ans-E
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 360-120=240
Re: Jamboree and GMAT Club Contest: How many words (with or without meanin &nbs [#permalink] 16 Nov 2015, 13:38

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