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December 6th, 2014, 08:57 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 09:25 AM. 
December 6th, 2014, 10:50 AM  #2  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, KingsOfLeon! Quote:
I assume we are forming 10digit strings of the digits. The first digit is 2to9: 8 choices. The last digit is 0to6: 7 choices. The 'middle' digits have $8!$ permutations. Answer: $\:(8)(8!)(7) \:=\: 2,257,920$  
December 6th, 2014, 11:59 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 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! 

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