Bunuel 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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Jeffery Miller

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