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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.
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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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!]
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!]
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 ?
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
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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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?
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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Think of the digits as the menu choices for a three-course 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 three-course meal possibilities = 2*9*4 = 72
It took me a little time to understand the question. Like, I had to read it thrice to formulate a concept in my head but once that was done, it was an easy ride.
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close
52% (01:31) correct 
