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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?
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 24 April 2014 - 12:51 AM
I'm sure I missed something but here is my best effort:
Posted 24 April 2014 - 05:17 PM
Spoiler for I'm going to venture a guess and say
Posted 26 April 2014 - 05:57 AM
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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