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Of the three-digit integers greater than 600, how many have [#permalink]
 11 Feb 2012, 17:04
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63% (00:53) wrong based on 2 sessions
Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two? (A) 120 (B) 116 (C) 108 (D) 107 (E) 72 I always struggle to solve these. What is the concept behind solving these questions?
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Re: How many integers? [#permalink]
11 Feb 2012, 17:10
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enigma123 wrote: Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 120
(B) 116
(C) 108
(D) 107
(E) 72
I always struggle to solve these. What is the concept behind solving these questions? # of three-digit integers greater than 600 is 399 (999-600); Numbers with all distinct digits: 4*9*8=288. First digit can take 4 values: 6, 7, 8 or 9, second digit can take 9 values (10 minus the one we used for the first digit) and third digit can tale 8 values (10 minus 2 digits we've already used for the first two digits); Numbers greater than 600 which have all alike digits: 4 (666, 777, 888, 999); {Total}-{all distinct}-{all alike}={two alike, one different} --> 399-288-4=107. Answer: D.
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Director
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Re: Of the three-digit integers greater than 600, how many have [#permalink]
11 Feb 2012, 17:22
Thanks for very clear and precise explanation Bunuel. Highly appreciate.
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Re: Of the three-digit integers greater than 600, how many have [#permalink]
04 Oct 2012, 14:05
If the problem asks for numbers greater than 600, don't we have to start counting from 601 forward? I think we don't, but just double-checking to make sure.... thanks!
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Re: Of the three-digit integers greater than 600, how many have [#permalink]
04 Oct 2012, 15:40
enigma123 wrote: Of the three-digit integers greater than 600, how many have two digits that are equal to each other and the remaining digit different from the other two?
(A) 120
(B) 116
(C) 108
(D) 107
(E) 72
I always struggle to solve these. What is the concept behind solving these questions? The question is about counting/combinations. The numbers can be of the form AAB, \,ABA, or ABB where A and B are different digits and A\geq6. There will be 4\cdot9\cdot3=108 possibilities, but we have to subtract 1 for the number 600 (which is obtained for A = 6 and B = 0). Therefore, total number of possibilities 108 - 1= 107.Answer D.
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Re: Of the three-digit integers greater than 600, how many have [#permalink]
29 Oct 2012, 16:04
Bunuel, thanks for the explanation. Mine looks more complicated, but could you please check if my solution is correct.
1) yxx - first digit can have 4 values, second - 9, third - 1 = 36
2) yyx - first digit - 4, second - 1, third - 9 = 36
3) yxy - first digit - 4, second 9, third - 1 = 36
1)+2)+3) = 108 - 1*= 107
1* - we should exclude 600
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Re: Of the three-digit integers greater than 600, how many have [#permalink]
30 Oct 2012, 14:28
Ev, what you did was basically the same as what evajager did. Let me clarify. Three digit number: 6xy First digit: either 6, 7, 8, or 9 So out of 4 possible digits, you choose 1. (4C1)Either 2nd or 3rd digit must match the first digit -> Only 1 possibility so (1C1)Then the remaining last digit (opposite of the you chose above) can have 9 remaining digits to choose from. So (9C1). Of course, out of 3 available spots, we have to pick 2 of them to be the same (3C2). So multiply them together: = # of ways to arrange a pair within 3 slots * [ digit #1 * pair digit * remaining digit]
= (3C2) [ (4C1) * (1C1) * (9C1) = 36 * 3 = 108Since the question technically asks for numbers >600, the exact value of 600 (which was included) should not be included. So we subtract 108 - 1 = 107
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Re: Of the three-digit integers greater than 600, how many have
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30 Oct 2012, 14:28
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