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 December 6th, 2014, 09:57 AM #1 Banned Camp   Joined: Dec 2014 From: Edinburgh Posts: 6 Thanks: 0 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 10:25 AM.
December 6th, 2014, 11:50 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

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

 December 6th, 2014, 12:59 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra 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! Thanks from topsquark

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