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# How many positive integers can be expressed as a product of

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Intern
Joined: 19 Jul 2008
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How many positive integers can be expressed as a product of [#permalink]

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25 Aug 2008, 19:19
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many positive integers can be expressed as a product of two or more of the primes numbers 5,7,11 and 13 if no one product is to include the same factor more than once?
A) 8 b) 9 c) 10 d) 11 e) 12

I got E as the answer ... don't have the OA to this one ... Please show your approach to solve and help to confirm ...

Kudos [?]: 12 [0], given: 0

Intern
Joined: 10 Aug 2008
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Location: Research Triangle Park, NC
Re: product of two or more of the primes numbers [#permalink]

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25 Aug 2008, 20:13
gmatcraze wrote:
How many positive integers can be expressed as a product of two or more of the primes numbers 5,7,11 and 13 if no one product is to include the same factor more than once?
A) 8 b) 9 c) 10 d) 11 e) 12

I got E as the answer ... don't have the OA to this one ... Please show your approach to solve and help to confirm ...

I think this is a simple case of permutations. So for this P(4,2) = 4!/(4-2)! = 24/2 = 12.

No?

cP
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Joined: 07 Nov 2007
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Location: New York
Re: product of two or more of the primes numbers [#permalink]

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25 Aug 2008, 20:58
gmatcraze wrote:
How many positive integers can be expressed as a product of two or more of the primes numbers 5,7,11 and 13 if no one product is to include the same factor more than once?
A) 8 b) 9 c) 10 d) 11 e) 12

I got E as the answer ... don't have the OA to this one ... Please show your approach to solve and help to confirm ...

= 4C2+4C3+4C4
= 6+4+1 =11

Will go for D.

How did you get 12.. Show your steps.
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Intern
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Location: Research Triangle Park, NC
Re: product of two or more of the primes numbers [#permalink]

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25 Aug 2008, 21:26
x2suresh wrote:
gmatcraze wrote:
How many positive integers can be expressed as a product of two or more of the primes numbers 5,7,11 and 13 if no one product is to include the same factor more than once?
A) 8 b) 9 c) 10 d) 11 e) 12

I got E as the answer ... don't have the OA to this one ... Please show your approach to solve and help to confirm ...

= 4C2+4C3+4C4
= 6+4+1 =11

Will go for D.

How did you get 12.. Show your steps.

x2suresh is right. This is a combinations problem not a permutations problem. -1 for me.

cP
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Joined: 16 Jul 2008
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Re: product of two or more of the primes numbers [#permalink]

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26 Aug 2008, 01:03
6 possible number with any two of the prime factors + 4 possible with three of the primes + 1 possible with all four = 11
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Re: product of two or more of the primes numbers   [#permalink] 26 Aug 2008, 01:03
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