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Bunuel
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Bunuel
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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Bunuel
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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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I did not quite understand the solution. Hi Bunuel - when figuring out the choices, why didn't we figure out the choices for the second digit first before the 3rd digit? For example,

In the range 900-999
1 choice for first digit: 9
9 choices for second digit: 10 possible digits, except the digit 9 which is the first digit already.
3 choices for third digit: 1,3,5,7,9, but subtract to account for the first and second digits.

Thank you!
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I did not quite understand the solution. Hi Bunuel - when figuring out the choices, why didn't we figure out the choices for the second digit first before the 3rd digit? For example,

In the range 900-999
1 choice for first digit: 9
9 choices for second digit: 10 possible digits, except the digit 9 which is the first digit already.
3 choices for third digit: 1,3,5,7,9, but subtract to account for the first and second digits.

Thank you!

That's not correct because if the second digit selected is even, you won’t have 3 options for the third digit—you’ll have 4. The general rule is to start with the slot that has the most restrictions and then move to the less restricted slots.
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