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Manager
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Permutation [#permalink] New post 03 Jul 2007, 06:32
In an examination a student can answer in 56 ways. If the number of questions asked exceeds the number of question to be answered by 2, then how many questions were to be answered?
A) 6
B) 8
C) 4
D) 2
E) 5
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Re: Permutation [#permalink] New post 03 Jul 2007, 06:57
pawan82 wrote:
In an examination a student can answer in 56 ways. If the number of questions asked exceeds the number of question to be answered by 2, then how many questions were to be answered?
A) 6
B) 8
C) 4
D) 2
E) 5


If the number of questions is x, then the required number can be answered in xP(x-2) ways.

Thus, xP(x-2) = 56
x = 8. (B)
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 [#permalink] New post 03 Jul 2007, 07:08
Sumande can you detail what is behind xP(x-2) please :oops:
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 [#permalink] New post 03 Jul 2007, 08:38
are you referring to binomial distribution... because i don't quite see where is the probability, but may be there is some more use of that formula??
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 [#permalink] New post 03 Jul 2007, 09:15
boubi wrote:
are you referring to binomial distribution... because i don't quite see where is the probability, but may be there is some more use of that formula??


P = permutation
There is no binomial distribution formula used!
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 [#permalink] New post 03 Jul 2007, 10:15
I remember seeing a similar question sometime back. The OE for that question was as follows

Let there be n questions. A person can answer a question in 2 ways by either answering it or not answering it.

Also we need to eliminate the possibility of not answering answering any of the questions. So the total number of ways to answer this whole test is (2^n) - 1. In that problem the number was 127. So 2^n -1 = 127 => n =7 But here the number is 56. Also Am not sure if permutation is the right way to go
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 [#permalink] New post 04 Jul 2007, 04:25
I think the order should matter, so permutation should be used.
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 [#permalink] New post 04 Jul 2007, 06:01
vshaunak@gmail.com wrote:
I think the order should matter, so permutation should be used.


That is what I thought. And, anyways, that was the only way I could get to a possible answer choice ! :lol:
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Re: Permutation [#permalink] New post 14 Sep 2007, 05:47
sumande wrote:
pawan82 wrote:
In an examination a student can answer in 56 ways. If the number of questions asked exceeds the number of question to be answered by 2, then how many questions were to be answered?
A) 6
B) 8
C) 4
D) 2
E) 5


If the number of questions is x, then the required number can be answered in xP(x-2) ways.

Thus, xP(x-2) = 56
x = 8. (B)


how are we working this out- xP(x-2) = 56?
Re: Permutation   [#permalink] 14 Sep 2007, 05:47
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