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December 6th, 2014, 08:57 AM   #1
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Question Is my answer correct

How many arrangements of the list 0, 1, 2, ... , 9.
First digit > 1
Last digit < 7


My answer:
10! - ( 10C8 * 10C7) = 3623400 arrangements.

Last edited by KingsOfLeon; December 6th, 2014 at 09:25 AM.
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December 6th, 2014, 10:50 AM   #2
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Hello, KingsOfLeon!

Quote:
How many arrangements of the list 0, 1, 2, ... , 9 ?
First digit > 1
Last digit < 7

I assume we are forming 10-digit strings of the digits.

The first digit is 2-to-9: 8 choices.
The last digit is 0-to-6: 7 choices.
The 'middle' digits have $8!$ permutations.

Answer: $\:(8)(8!)(7) \:=\: 2,257,920$

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December 6th, 2014, 11:59 AM   #3
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I don't agree with that analysis. If the first digit is less than 7, your count for the last digit is wrong.

I would count all permutations, subtract those that begin with 0 or 1 and those that end with 7, 8 or 9, and then add back those that both begin with 0 or 1 and end with 7, 8 or 9.

10! - 2 x 9! - 3 x 9! + 2 x 3 x 8!
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