Kenneth Rosen Edition 6 Question 45 (Page No. 346)

Question

How many bit strings of length eight contain either three consecutive 0s or four consecutive 1s?

Comments

@Arjun

sir, can you check these answers ?

and i want to know, is there any other approach to this question ?

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This is one of the lengthy question. I still got different solution with different approach. Plz correct me if something missing

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@ Shaikh bro superbb explaination

Answer 1

I have found Number of bit string with 

(1) (i)zero consecutive $0.$
(ii) one $0$
(iii) two  consecutive zero

or

(2)(i) zero consecutive 1s
(ii) one $1$
(iii) two consecutive $1's$
(iv)three consecutive $1's$

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Now,

for (1)(i) I got $1$ string
 
(ii)$4$ string eliminated duplicates

(iii) $7+5+3+1=16$ string

(iv) we can arrange like (2,1) in two places or (1,1,1) in $3$ places

In 1st one I got 4+3+2+1=10 strings

In 2nd one I got 4+2+(5*4/2)=12 strings

Now we can take four $0's$ and arrangement will be (2,2), so that no three consecutive 

Here I got 4+2=6 strings

Again I can take five consecutive $0's$ and arrangement will be like (2,2,1)

Here I got 2 strings

And lastly I can take six $0's$, arrangement will be like (2,2,2), here I got $1$ string

So, total 52 strings
 

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Now for $1's$

2)(i) that is string with no $1's=1$ string

(ii) string with one $1's$=4 string

(iii) No. of string with three $1's$

Arrangement can be (1,1,1), and there are $6+5+4=15$ strings

Arrangement can be (1,2) and there will be 4+3=7 strings

(:)Now  if there are four $1's$ and their arrangement will be (3,1) or arrangement (2,2)

there will be (9+7)=16 strings

(:) If there are five $1's$ arrangement will be like (3,2) or (2,2,1)

There will be 6+3=9 strings

(:) If there are  six 1's arrangement will be (2,2,2) or (3,3) 

There will be $16$ strings

Total $68$ strings

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Now total string-(All previous string ) coming 136