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Mathematics P6 NOV 2011

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Don't know...
Lemme explain my reasoning.
Two EE's together-EE can be considered as one letter only because order doesn't matter here, both letters are E itself.
That leaves me with X T R E M P A P R S EE
E is not same as EE, so I have 11 letters in all, out of which there are 2 P's.
So i got 11!/2! .
Well you see, I missed the fact that there are 2 R's.
So, final answer=11!/2!2!
Yeah you're right. Thanks shressubha and Emily793!!

Anyway i don't really care abt stats now I screwed the paper :fool: I missed the full marks!!!!
 
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Hmm... why not 11!/ (2!2!2!)?

"Two EE's together-EE can be considered as one letter only because order doesn't matter here, both letters are E itself."

So, there are 2Es, 2Rs and 2 Ps.
 
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There are 3 Es. I'm considering only 2 of them as one whole => there are 2Es; one E and one 'EE'.
 
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auragirl said:
I completely forgot my answers now.
You don't put them in a recycle bin? It would be much appreciated if you could simply move it back. :)

If I could... I would have... :)
 
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Emily793 said:
heys guys how were we supposed to do ques 2?? the head and tail thing??

I think 2^12...
I listed and calculated all possibilities using 3 coins instead of 12. I got 8. So i figured that would be 2^3=8. Since it was 12 in the question i just substituted 3 by 12 and ended up with 2^12. I'm not sure but it's the best logic I've come up with, like the closest I can get to a sensible answer.
 
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CaptainDanger said:
auragirl said:
I completely forgot my answers now.
You don't put them in a recycle bin? It would be much appreciated if you could simply move it back. :)

If I could... I would have... :)

No worries. :)

Anahita, the Pigeon Hole method. It's good. I used the Factorial method, you get the same answer. :)
 
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[12!/11! + 12!/2!10! + 12!/3!9! + 12!/4!8! + 12!/5!7!]*2 +12!/6!6!
Got it right after exams.
 
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but even after that te answer is different to 2^12 right so hows this coorect and why 2 times in the 12!/5!x7!
 
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se when you do the way you did you get 3301 as the answer whereas if you 2^12 you get 4096 and by the way why do you multiply 2 times to 12!/7!x5!
 
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You get the same answer. -.-

I multiplied all of these by 2:
12!/11! + 12!/2!10! + 12!/3!9! + 12!/4!8! + 12!/5!7!

There's a bracket there.
And please, it's over. You've other papers.
 
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Yeah, but you missed, I missed it, doesn't mean others missed it too.^

Anyway, stale paper now. Forget it. Prepare for the ones to come. :)

Thanks Anahita. :)
 
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