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If each of the digits is to be used only once, what is the

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CEO
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If each of the digits is to be used only once, what is the [#permalink]

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New post 25 Feb 2008, 11:32
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If each of the digits is to be used only once, what is the probability of creating a 3-digit number that is divisible by ten?
(A)9/125
(B) 81/1000
(C) 1/10
(D) 18/25
(E) 729/1000
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You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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Re: Counting [#permalink]

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New post 25 Feb 2008, 12:46
bmwhype2 wrote:
If each of the digits is to be used only once, what is the probability of creating a 3-digit number that is divisible by ten?
(A)9/125
(B) 81/1000
(C) 1/10
(D) 18/25
(E) 729/1000


seems answer choises are wired.

= (9 x 8 x 1) / (900)
= 2/25

so close is 1/10 but its not correct.

if 0 can also be used in hundreds place, then it is:

= (10x9x1)/(9x10x10)
= 1/10
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CEO
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Kudos [?]: 1076 [0], given: 4

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Re: Counting [#permalink]

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New post 25 Feb 2008, 12:51
I also thought the question was weird.

Zero cannot be the 1st digit in a number and only C makes sense if we are allowed to have zero in the hundredths slot,


The OA is A.
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You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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Manager
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Re: Counting [#permalink]

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New post 25 Feb 2008, 18:40
i am getting 1/10 also.
do they have explanation in the OA?

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CEO
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Re: Counting [#permalink]

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New post 25 Feb 2008, 19:07
bmwhype2 wrote:
If each of the digits is to be used only once, what is the probability of creating a 3-digit number that is divisible by ten?
(A)9/125
(B) 81/1000
(C) 1/10
(D) 18/25
(E) 729/1000



hrmmmmm I get 9*8*1/(9*10*10) --> 2/25

???

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Re: Counting   [#permalink] 25 Feb 2008, 19:07
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