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How many 3-digit integers can be chosen such that

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How many 3-digit integers can be chosen such that [#permalink] New post 22 Aug 2017, 02:26
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How many 3-digit integers can be chosen such that none of the digits appear more than twice, and none of the digits equal 0?

(A) 729
(B) 720
(C) 648
(D) 640
(E) 576
[Reveal] Spoiler: OA

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Re: How many 3-digit integers can be chosen such that [#permalink] New post 25 Aug 2017, 04:16
Can somebody explain? I think I did not get the question.
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Re: How many 3-digit integers can be chosen such that [#permalink] New post 25 Aug 2017, 06:09
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boxing506 wrote:
Can somebody explain? I think I did not get the question.


The question askes you to select a 3 digit number. So we need to selct 3 digits from {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Now number of ways we can select 1 digits = 9. Since there are 9 options to choose from!
Now number of ways we can select 3 digits = \(9 \times 9 \times 9 = 729\)

Now the above section contains digits like = 999, 888, 777 ..... they are not allowed (3-digit integers with all alike digits )


So correct answer is \(729 - 9 = 720\). Hence B.
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Last edited by Carcass on 25 Aug 2017, 06:47, edited 1 time in total.
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Re: How many 3-digit integers can be chosen such that [#permalink] New post 16 May 2018, 09:24
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Carcass wrote:


How many 3-digit integers can be chosen such that none of the digits appear more than twice, and none of the digits equal 0?

(A) 729
(B) 720
(C) 648
(D) 640
(E) 576


Let’s first disregard the condition of none of the digits appearing more than twice and count the three-digit numbers where none of the digits is 0. We see that there are 9 choices for each of the digits; therefore, there are 9^3 = 729 such numbers.

Now, we can deal with the condition that none of the digits should appear more than twice. If a digit of a 3-digit number appears more than twice, then it must appear all 3 times and there are only 9 numbers that have this property: 111, 222, …, 999. Thus, out of the 729 numbers, 9 do not satisfy this property and 729 - 9 = 720 do satisfy.

Answer: B
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Re: How many 3-digit integers can be chosen such that [#permalink] New post 17 Apr 2019, 00:02
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The other solution is

3 digit numbers that have at most 2 same digits and 1 different are 9*1*8*3=216.
3digit numbers whose all digits are different are 9*8*7=504

Adding 504+216 gives us 720 which is B.
Re: How many 3-digit integers can be chosen such that   [#permalink] 17 Apr 2019, 00:02
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