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
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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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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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Amazingly elegant method.
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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!]

Check other ways of solving HERE.

Hope it helps.
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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 ?

I don't agree with the explanation. Why can't 9 be an option for the first digit?

Please re-read the solution above. We do consider 9 as the hundreds digit there.
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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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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.

You can also split the answer by cases: 1) 800s: 8(odd)(odd)=5*4, 8(even)(odd)=4*5, etc.
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Hi,

I did this solution in a different manner.

Total number of 3 digit numbers with different digits >800 = 2 (for 8 or 9) * 9 (for all digits except the first one) * 8 (for all digits except 1st and 2nd) = 144

Since every number can either be an odd or an even, exactly half of these 144 should be odd.

Hence, solution = 72.

Regards.

Total 3 digits numbers greater than 800 with 3 different digits = 9*8+9*8 =144
Half will be even half will be odd
Hence 144/2 = 72

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