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Computer Science 11 Online
OpenStudy (anonymous):

How many 10 bit binary strings are there none of which contains the pattern '110'.

OpenStudy (anonymous):

the subtraction part:

OpenStudy (anonymous):

Hmm...

OpenStudy (anonymous):

let's count all the strings which have one 110: you realize that there are (2^7)*7 of them,right?

OpenStudy (anonymous):

Yeah, right

OpenStudy (anonymous):

So there must be a total of 2^10-(2^7)*7 strings which does not contain '110'. I think i have got it.

OpenStudy (anonymous):

NO NO NO

OpenStudy (anonymous):

that's what i was trying to say. If you only discount the strings which have one 110, you'll have left out on strings that contain 2 or 3 110s. Think about it.

OpenStudy (anonymous):

Oh yes. You are right. I did not consider that condition.

OpenStudy (anonymous):

I guess no.

OpenStudy (anonymous):

ok np i'll explain

OpenStudy (anonymous):

Thank You man. You are very helpful. :D

OpenStudy (anonymous):

Okay yes, I got it.It should be 2^4 *4

OpenStudy (anonymous):

??

OpenStudy (anonymous):

Aww thanks :) so, you have 2 110s. the configurations possible are: (strings where there's no gap between the 110s) 110110____ _110110___ __110110__ ___110110_ ____110110 (strings where there's one gap between the 110s) 110_110___ _110_110__ __110_110_ ___110_110 (strings where there's two gaps between the 110s) 110__110__ _110__110_ __110__110 (strings where there's three gaps between the 110s) 110___110_ _110___110. That makes 14 configurations. With each of these configurations, you can make 2^4 numbers. So, the total number of strings with 2 110s are 14(2^4). Do you get it?

OpenStudy (anonymous):

disregard that (2^4)*4 reply. that's wrong.

OpenStudy (anonymous):

Thank You for your help.!

OpenStudy (anonymous):

It's still not done though :\

Parth (parthkohli):

=/ @thenewguy: You do realize that I explained it to you.

Parth (parthkohli):

Sorry I was wrong btw

OpenStudy (anonymous):

you weren't *wrong*. Just.....incomplete. I realized it after you went offline yesterday.

Parth (parthkohli):

I am silly at combinatorics :|

OpenStudy (anonymous):

Actually i am sorry that i couldn't just come up online. There was some problem with my internet connection, so I had to go back offline. I am really thankful to both of you for your help. Can u guys please just tell me what the final answer should be.? Thank You!

OpenStudy (anonymous):

@ParthKohli: Yes, offcourse. I am really thankful for your help. :)

OpenStudy (konradzuse):

Did we get this answer? Such a stupid Q imo... More of a math problem than comp sci...

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