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November 19th, 2011, 03:26 AM   #1
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permutation



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November 19th, 2011, 06:55 AM   #2
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Re: permutation

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November 19th, 2011, 07:42 AM   #3
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Re: permutation



Assuming first digit can be zero.
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November 19th, 2011, 07:59 AM   #4
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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
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November 19th, 2011, 09:44 AM   #5
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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.
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November 19th, 2011, 10:45 AM   #6
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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.


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[color=beige]. . [/color]






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November 19th, 2011, 10:56 AM   #7
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Re: permutation

Quote:
Originally Posted by Hoempa
Hoempa,

It should be

And then you have the same result as soroban and mrtwhs.
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November 20th, 2011, 03:28 AM   #8
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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.
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November 21st, 2011, 11:11 PM   #9
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Re: permutation

Thanks to all for nice explanation.

answer given is =5292
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