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December 27th, 2017, 04:41 AM  #1 
Member Joined: Dec 2017 From: Tel Aviv Posts: 44 Thanks: 3  The number of choices
What is the number of choices to compose a number with the 3 digits: 1, 3, 9? (The digit need no to be repeated.) What is the formula? I forget it, so if you can write it, it would be helpful. Thanks. Last edited by skipjack; December 27th, 2017 at 09:15 AM. 
December 27th, 2017, 05:02 AM  #2 
Newbie Joined: Nov 2017 From: US Posts: 5 Thanks: 4 
As the digits cannot repeat, the first digit can be either 1, 3 or 9 (3 options). Second digit can be one of the remaining 2 (2 options) and the final digit is the left over one (1 option). Hence total number of choices you have is 3*2*1 = 6. 139 193 319 391 913 931 
December 27th, 2017, 05:42 AM  #3 
Member Joined: Dec 2017 From: Tel Aviv Posts: 44 Thanks: 3 
Nice. Thanks... it is n!, I think... Last edited by policer; December 27th, 2017 at 05:49 AM. 
December 27th, 2017, 08:10 AM  #4 
Newbie Joined: Nov 2017 From: US Posts: 5 Thanks: 4 
Yes, you got it!.

December 27th, 2017, 08:47 AM  #5 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,966 Thanks: 807 
Saying the "digits need not be repeated" (as in #1) is not the same as saying "the digits cannot be repeated" (as in #2).

December 27th, 2017, 09:19 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,704 Thanks: 1529 
The wording in post #1 is "need no to be", which is closer to "need not to be" than "need not be".


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