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Re M2520 [#permalink]
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Official Solution:How many odd threedigit integers greater than 800 are there such that all their digits are different? A. 40 B. 60 C. 72 D. 81 E. 104 In the range 800  900: 1 choice for the first digit: 8 5 choices for the third digit: 1, 3, 5, 7, 9 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: \(1*5*8 = 40\). In the range 900  999: 1 choice for the first digit: 9 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's the first digit) 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: \(1*4*8 = 32\). Total: \(40+32 = 72\). Answer: C
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Re: M2520 [#permalink]
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28 Feb 2015, 12:23
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Bunuel wrote: Official Solution:
How many odd threedigit integers greater than 800 are there such that all their digits are different?
A. 40 B. 60 C. 72 D. 81 E. 104
In the range 800  900: 1 choice for the first digit: 8 5 choices for the third digit: 1, 3, 5, 7, 9 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: \(1*5*8 = 40\). In the range 900  999: 1 choice for the first digit: 9 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's the first digit) 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: \(1*4*8 = 32\). Total: \(40+32 = 72\).
Answer: C Hi Bunuel, I just wanted to clarify this question, as i didn't understand it properly. If the question is asking how many three digit odd integers then why are we including the numbers from 801  899...shouldn't these numbers not be counted because these numbers have an 8 in them? (even number) . I thought the question is asking the possibilities of 3 digit odd integers that are all different..example 913  all odd and the digits are different. I hope my question makes sense, i'm just a bit confused. Thank you.



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Re: M2520 [#permalink]
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01 Mar 2015, 04:30
pinkcupcake wrote: Bunuel wrote: Official Solution:
How many odd threedigit integers greater than 800 are there such that all their digits are different?
A. 40 B. 60 C. 72 D. 81 E. 104
In the range 800  900: 1 choice for the first digit: 8 5 choices for the third digit: 1, 3, 5, 7, 9 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: \(1*5*8 = 40\). In the range 900  999: 1 choice for the first digit: 9 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's the first digit) 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: \(1*4*8 = 32\). Total: \(40+32 = 72\).
Answer: C Hi Bunuel, I just wanted to clarify this question, as i didn't understand it properly. If the question is asking how many three digit odd integers then why are we including the numbers from 801  899...shouldn't these numbers not be counted because these numbers have an 8 in them? (even number) . I thought the question is asking the possibilities of 3 digit odd integers that are all different..example 913  all odd and the digits are different. I hope my question makes sense, i'm just a bit confused. Thank you. 801, 803, 805, 807, ... are all odd integers.
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Re: M2520 [#permalink]
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01 Mar 2015, 18:54
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Bunuel wrote: Official Solution:
How many odd threedigit integers greater than 800 are there such that all their digits are different?
A. 40 B. 60 C. 72 D. 81 E. 104
In the range 800  900: 1 choice for the first digit: 8 5 choices for the third digit: 1, 3, 5, 7, 9 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: \(1*5*8 = 40\). In the range 900  999: 1 choice for the first digit: 9 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's the first digit) 8 choices for the second digit: 10 digits  first digit  third digit = 8 digits. Total for this range: \(1*4*8 = 32\). Total: \(40+32 = 72\).
Answer: C Amazingly elegant method.
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Re: M2520 [#permalink]
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25 May 2015, 07:26
Thanks for your explanation! Is there an alternative method to solve this question? trying to think how could this problem be solve by applying the combinatorics formula with repeating elements [N!][/A!B!]



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25 May 2015, 07:31



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Re M2520 [#permalink]
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04 Nov 2015, 16:03
I think this is a highquality question and the explanation isn't clear enough, please elaborate. I don't get explanation for the second digit part can you please list out all possibilities ?



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Re M2520 [#permalink]
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29 Jun 2016, 12:01
I don't agree with the explanation. Why can't 9 be an option for the first digit?



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Re: M2520 [#permalink]
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11 Nov 2016, 14:42
3 digit integers such that all three digits are different between 800899 : 100th place is reserved for 8, 10th place can be taken by any of the remaining 9 digits (0,1,2,3,4,5,6,7,9) Unit's pace can be taken by any of the remaining 8 digits so 9*8 = 72 such numbers Similary there are 72 such numbers between 900999 Total number 72 + 72 =144 Between 800999 (both inclusive) there are 200 numbers, so there should be equal number of odd and even numbers: hence total odd numbers = 144/2 = 72



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Re: M2520 [#permalink]
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01 Apr 2017, 10:12
I dont get the explanation for the 2nd digit. 800899 1st digit: 1 choice (8) 3rd digit: 5 choices (1,3,5,7,9) 2nd digit: 10 digits  1 digit  5 digits (0,2,4,6)
This way, you cannot get 811. If I use your solution above, it does not exclude 811, 822, 833, 844 etc.



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Re: M2520 [#permalink]
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01 Apr 2017, 10:18
Kontaxis wrote: I dont get the explanation for the 2nd digit. 800899 1st digit: 1 choice (8) 3rd digit: 5 choices (1,3,5,7,9) 2nd digit: 10 digits  1 digit  5 digits (0,2,4,6)
This way, you cannot get 811. If I use your solution above, it does not exclude 811, 822, 833, 844 etc. The first step when solving a question ALWAYS must be reading CAREFULLY: How many odd threedigit integers greater than 800 are there such that all their digits are different?
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Re: M2520 [#permalink]
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02 Apr 2017, 05:19
Bunuel wrote: The first step when solving a question ALWAYS must be reading CAREFULLY: How many odd threedigit integers greater than 800 are there such that all their digits are different? Hi Bunuel, I understand the question stem to explicitly exclude cases where some of the 3 digits are the same, i.e. numbers such as 811, 822, 833 should be excluded. That is why I have less choices for the 2nd digit in my solution as you have. That way I ensure that none of the 3 digits are the same. Could you please elaborate on this?



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Re: M2520 [#permalink]
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02 Apr 2017, 05:28
Kontaxis wrote: Bunuel wrote: The first step when solving a question ALWAYS must be reading CAREFULLY: How many odd threedigit integers greater than 800 are there such that all their digits are different? Hi Bunuel, I understand the question stem to explicitly exclude cases where some of the 3 digits are the same, i.e. numbers such as 811, 822, 833 should be excluded. That is why I have less choices for the 2nd digit in my solution as you have. That way I ensure that none of the 3 digits are the same. Could you please elaborate on this? So, you say that it should be  In the range 800  900: 1*5*4? That's not right. For the 2nd digit you can use any digit apart from those used for the first and the thrid one, so the number of options is 10  1  1 = 8. If still not convinced check other ways of solving HERE.
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Re: M2520 [#permalink]
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08 Sep 2017, 15:01
Think of the digits as the menu choices for a threecourse meal: Appetizers (hundreds digit), Main (tens digit), and Desserts (units digit)
Appetizer menu has 2 choices: {8,9}  eat "9" Main menu has 3 choices: {0,1,2,3,4,5,6,7,8} i.e. "9" is already eaten Dessert menu has 4 choices: {1,3,5,7} i.e. again, "9" is already eaten
Total threecourse meal possibilities = 2*9*4 = 72










