First count number of $3$ consecutive "0" in $10$ bit binary string.$000$ can start at position $1,2,3,4,5,6,7,8$
$\begin{vmatrix} 0 &0 &0 &X &X &X &X &X &X &X \\ 1&0 &0 &0 &X &X &X &X &X &X \\ X&1 &0 &0 &0 &X &X &X &X &X \\ X& X &1 &0 &0 &0 &X &X &X &X \\ X& X &X &1 &0 &0 &0 &X &X &X \\ X& X &X &X &1 &0 &0 &0 &X &X \\ X&X &X &X &X &1 &0 &0 &0 &X \\ X& X &X &X &X &X &1 &0 &0 &0 \end{vmatrix}$
Now,Total combination=$2^{7}+6*2^{6}=512$
But,we have counted many string twice,so we need to subtract from the final combination
How to find the repeated String?
Let we have stored the possible string in array$ar\left [ i \right ]\left [ j \right ],1\leq i\leq 8 , 1\leq j\leq 10$
for $ar\left [ 1 \right ]\left [ X \right ]$,no strings are repeated with any of $ar\left [ X \right ]\left [ X \right ]$
(take a string from a row and compare to other row )
for $ar\left [ 2 \right ]\left [ X \right ]$ try to match with $ar\left [ 6 \right ]\left [ X \right ]$,$ar\left [ 7 \right ]\left [ X \right ]$,$ar\left [ 8 \right ]\left [ X \right ]$
The matched string will be $\Rightarrow 10001000XX$(4 possible string),$1000X1000X$(4 possible string) ,$1000XX1000$
Do same for $ar\left [ 3 \right ]\left [ X \right ]$,repeated strings are $X10001000X$(4 possible String),$X1000X1000$( 4 possible string)
and the last for $ar\left [ 4 \right ]\left [ X \right ]$,$XX10001000$(4 possible string)
So Total number of 3 consecutive string =512-(4*6)=488
Do exactly same for 3 consecutive 1
Number of 10 bit binary string where the string contains 3 consecutive 0's or 3 consecutive 1's =488+488=976
Answer=966
