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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Guest Message by DevFuse

Started by BMAD, Apr 22 2014 03:12 AM

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

13 replies to this topic

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.

Posted 23 April 2014 - 02:00 PM

My guess is that it is a pretty big number!

*[sorry I couldn't pass that up]*

Posted 23 April 2014 - 03:10 PM

Posted 23 April 2014 - 05:48 PM

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

Posted 23 April 2014 - 08:13 PM

because?

Posted 24 April 2014 - 12:51 AM

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

Spoiler for

Posted 24 April 2014 - 01:58 AM

Spoiler for Still struggling with it but

Posted 24 April 2014 - 05:17 PM

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

because?

Spoiler for I did

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!)

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!)

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

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