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Re M25-20 [#permalink]
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 16 Sep 2014, 01:23
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Official Solution:How many odd three-digit 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: M25-20 [#permalink]
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28 Feb 2015, 12:23
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Bunuel wrote: Official Solution:
How many odd three-digit 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: M25-20 [#permalink]
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01 Mar 2015, 04:30
pinkcupcake wrote: Bunuel wrote: Official Solution:
How many odd three-digit 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: M25-20 [#permalink]
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01 Mar 2015, 18:54
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Bunuel wrote: Official Solution:
How many odd three-digit 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: M25-20 [#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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Re: M25-20 [#permalink]
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25 May 2015, 07:31
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Re M25-20 [#permalink]
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04 Nov 2015, 16:03
I think this is a high-quality 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 M25-20 [#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: M25-20 [#permalink]
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30 Jun 2016, 08:01
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Re: M25-20 [#permalink]
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11 Nov 2016, 14:42
3 digit integers such that all three digits are different between 800-899 :
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 900-999
Total number 72 + 72 =144
Between 800-999 (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: M25-20 [#permalink]
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01 Apr 2017, 10:12
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I dont get the explanation for the 2nd digit.
800-899
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: M25-20 [#permalink]
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01 Apr 2017, 10:18
Kontaxis wrote: I dont get the explanation for the 2nd digit.
800-899
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 three-digit integers greater than 800 are there such that all their digits are different?
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Re: M25-20 [#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 three-digit 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: M25-20 [#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 three-digit 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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