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

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CEO
Joined: 21 Jan 2007
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If each of the digits is to be used only once, what is the [#permalink]  25 Feb 2008, 10:32
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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Joined: 29 Aug 2007
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Re: Counting [#permalink]  25 Feb 2008, 11:46
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

= (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
_________________
CEO
Joined: 21 Jan 2007
Posts: 2764
Location: New York City
Followers: 9

Kudos [?]: 348 [0], given: 4

Re: Counting [#permalink]  25 Feb 2008, 11:51
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

Manager
Joined: 12 Feb 2008
Posts: 181
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Re: Counting [#permalink]  25 Feb 2008, 17:40
i am getting 1/10 also.
do they have explanation in the OA?
CEO
Joined: 29 Mar 2007
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Re: Counting [#permalink]  25 Feb 2008, 18:07
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

???
Re: Counting   [#permalink] 25 Feb 2008, 18:07
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