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Re: How many 3-digit integers can be chosen such that [#permalink]
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



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]
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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.
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Re: How many 3-digit integers can be chosen such that [#permalink]
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