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# Unique Codes

Best Answer m00li, 27 April 2014 - 01:05 PM

Spoiler for Still struggling with it but

EDIT: the last term for F(n) should read n-[n/2]C[n/2]f(n-[n/2]) where [k] is the max integer <=k

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Posted 22 April 2014 - 03:12 AM

Make a 10 digit code using the digits 0-9 each once.  Make another 10-digit code without any three adjacent code sequences repeating.  How many unique codes can be made following these two rules?

For example:

code 1: 0,1,2,3,4,5,6,7,8,9

code 2 cannot contain 0,1,2 or 1,2,3 or 2,3,4 or ,3,4,5, etc. anywhere in its code.

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### #2 dgreening

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Posted 23 April 2014 - 02:00 PM

My guess is that it is a pretty big number!

[sorry I couldn't pass that up]

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Posted 23 April 2014 - 03:10 PM

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### #4 Rob_G

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Posted 23 April 2014 - 05:48 PM

Spoiler for I'm going to venture a guess and say

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Posted 23 April 2014 - 08:13 PM

because?

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### #6 Dr XP

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Posted 24 April 2014 - 12:51 AM

I'm sure I missed something but here is my best effort:

Spoiler for

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### #7 m00li

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Posted 24 April 2014 - 01:58 AM

Spoiler for Still struggling with it but

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### #8 Rob_G

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Posted 24 April 2014 - 05:17 PM

Spoiler for I'm going to venture a guess and say

because?

Spoiler for I did

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### #9 carefreekaran

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Posted 26 April 2014 - 05:57 AM

Total cases = 10!.
Considering case when one block( 3 digits) is same as 1st code sequence= 8!..
Multiply by 2 because 3 digits can be reversed in a block and they'll still be adjacent..
Total blocks= 8. Hence total cases=8*2*8!..
[Above expression also contains cases when more than 3 digits are same w.r.t 1st code..]
Therefore ans= 10!-(2*8*8!)
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### #10 m00li

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Posted 27 April 2014 - 01:05 PM   Best Answer

Spoiler for Still struggling with it but

EDIT: the last term for F(n) should read n-[n/2]C[n/2]f(n-[n/2]) where [k] is the max integer <=k

• 0

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