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M32-09

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M32-09 [#permalink]

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Certain word is written on a paper. What is the number of arrangements of letters of that word ?



(1) If the first two letters were omitted, the number of arrangements of letters of shortened word would be 6

(2) There are 5 letters in the word
[Reveal] Spoiler: OA

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Official Solution:


Certain word is written on a paper. What is the number of arrangements of letters of that word ?

(1) If the first two letters were omitted, the number of arrangements of letters of shortened word would be 6. This one is clearly insufficient: we don't know how many letters are repeated in the word or even how many letters are there. Not sufficient.

(2) There are 5 letters in the word. We don't know how many letters are repeated in the word. Not sufficient.

(1)+(2) We can deduce that the last three letters of the word are all different (hence their arrangement of \(3! = 6\)) but we still don't know whether they repeat any of the first two letters. For example, if the word is goose, then the number of arrangements of its letters would be \(\frac{5!}{2!}\) but if the word is close, then the number of arrangements of its letters would be \(5!\). Not sufficient.


Answer: E
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Re: M32-09 [#permalink]

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Bunuel wrote:
Certain word is written on a paper. What is the number of arrangements of letters of that word ?



(1) If the first two letters were omitted, the number of arrangements of letters of shortened word would be 6

(2) There are 5 letters in the word



Ofcourse solution remains as above..
Just a few examples included...

1) two letters omitted, arrangements become 6..
One case... Remaining all are different... 3!=6... Total letters 3+2
Second case.. few remaining are common... 4 letters with 2 of one kind and other two of similar kind...\(\frac{4!}{2!/2!}\)=6
Here elements are 4+2

So there can be different elements and we don't even know if the two removed are same or different.

Insufficient

2) there are five letters.
All five could be same 5!/5!=1
All different=5!
Insufficient

Combined..
5 letters..
2 removed, remaining 3 in 6 arrangements... So SURELY these 3 are different letters.
But WHAT about the 2 removed..
Both same but different from the three left.. 5!/2!
Both same as one already in 3... 5!/3!
All different..5!
And so on
Insufficient

E
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Re M32-09 [#permalink]

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New post 22 Sep 2017, 21:21
I think this is a high-quality question and I agree with explanation.

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Re M32-09   [#permalink] 22 Sep 2017, 21:21
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