November 19th, 2011, 03:26 AM  #1 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16  permutation 
November 19th, 2011, 06:55 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond  Re: permutation 
November 19th, 2011, 07:42 AM  #3 
Senior Member Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0  Re: permutation Assuming first digit can be zero. 
November 19th, 2011, 07:59 AM  #4 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: permutation ways to pick 5 different digits out of 10 (0 to 9) They can be arranged in 5! ways. one in five numbers will have the middle digit largest. Would yield However You might exclude numbers with 0 as leading digit. of these with different digits. Leaving only 
November 19th, 2011, 09:44 AM  #5 
Senior Member Joined: Feb 2010 Posts: 707 Thanks: 142  Re: permutation
Well I'm getting a different answer. I'm assuming that the first digit is not zero and that the middle digit (the largest) must be 4,5,6,7,8, or 9. 
November 19th, 2011, 10:45 AM  #6  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: permutation Hello, panky! I agree with mrtwhs. Quote:
I will assume that the number must not begin with zero. There are two cases to consider. [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color] [color=beige]. . [/color]  
November 19th, 2011, 10:56 AM  #7  
Senior Member Joined: Oct 2011 From: Belgium Posts: 522 Thanks: 0  Re: permutation Quote:
It should be And then you have the same result as soroban and mrtwhs.  
November 20th, 2011, 03:28 AM  #8 
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: permutation Thanks, wnvl! So it is like: if 0 is the leading digit, there are 9 choices for the 4 digits, which can be arranged in 4! ways. In 1 in 4 cases, the 2nd of the 4 (middle digit of the no.) will be the largest. So possibilities. 
November 21st, 2011, 11:11 PM  #9 
Senior Member Joined: Jul 2011 Posts: 405 Thanks: 16  Re: permutation
Thanks to all for nice explanation. answer given is =5292 

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