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salama
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tapan22
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salama
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fankor
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let me start with a simple example.
if 5!/4!, then answer is 5 because (5*(4!))/4!.
Move onto a bit complicated one.
if (6!-5!)/4!, then (6*5!-5!)/4! -> here, the common factor in the numerator is 5!.
So 5!(6-1)/4!.
Then (5!*5)/4! -> change 5! to 5*4! -> (5*4!*5)/4! -> eliminate 4! -> answer is 25.
Apply the same logic, and you will understand this kind of factorial problem.
Let me know if I am wrong.
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tapan22
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salama
Cheers that really helps is it possible for you to elaborate on this step
= 64! (67*66*65 - 65)/64!

So you are saying that 64! is the common term. Any feedback will be appreciated.


(67! - 65!)/64!

= ((67*66*65*64!) - (65*64!))/64! --->( 67! = 67*66*65*64*...which is 64!)

From here just factor out 64! form the numerator.
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salama
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Tapan22 and Fankor.. you guys are awesome thanks for the detailed explanations...
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jaynayak
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(67! -65!)/64!
= (67*66*65! -65!)/64!
= 65!(67*66 - 1) 64!
= 65*64!(67*66 -1)/64!
= 65(67*66 -1)
= 287365
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GMATPsycho
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I agree with all..... but what if the answer choices are in factorials????
For example, Jaynayak's calculation was exactly what I followed that resulted an answer 287365 (This would have been a little tougher for me without the calculator, however...), my question here is what if the answer choices were something like 68!, 67!, 64!, 63!..... What then???
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(67! - 65!)/64!
= (67.66.65.64! - 65.64!)/64!
= 65.64!.(67.66-1)/64!
= 65(67.66 -1)

I don't think converting this back into factorial is possible.
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ywilfred
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Take out common factors !!

67! = (67)(66)(65)(64)(63)....
65! = (65)(64)(63)....

67! - 65!
= (67)(66)(65)(64)(63).... - (65)(64)(63)....
= (65!) (67*66 - 1)
= 65(64!)(4421)

So 67! - 65!/64! = 65(4421) = 287365



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