Let $a_{n}$ be the number of bit strings containing three consecutive zeroes.
Then if first bit is 1 we can choose 3 zeroes in $a_{n-1}$ ways .
or if it is 01 we can choose 3 zeroes in $a_{n-2}$ ways.
or if it is 001 then we can choose 3 zeroes in $a_{n-3}$ ways.
and we have exactly one set of 3 zeroes with 3 digits so $2^{n-3}$
$a_{n} = a_{n-1} + a_{n-2} + a_{n-3} + 2^{n - 3}$
$a_{0}=a_{1}=a_{2}=0,a_{3}=1$
$a_{4} = a_{3} + a_{2} + a_{1} + 2^1$ = 3
$a_{5} = a_{4} + a_{3} + a_{2} + 2^{ 2 }$ = 8
$a_{6}=20 , a_{7} = 47, a_{8}=107,a_{9}=238, a_{10}=520 $
Similarly for 3 consecutive 1's we get 520 ways.
Now we find intersection of both 000 and 111. Remaining 4 spaces can be filled in $2^4 = 16$ ways.
000 can occur in a string of length 10 in eight ways - from positions 1 to 8. Lets consider them separately and assume 000 comes first (to make the cases exhaustive but still mutually exclusive).
- 000 _ _ _ _ _ _ _ - 5 places for 111 to come - 5*16 possible strings as remaining 4 places can be filled in 16 ways
- 1 000 _ _ _ _ _ _ - 4 places for 111 to come - 4*8 possible strings as remaining 3 places can be filled in 8 ways (1 is added before 000 as 0 coming in first position is counted in part 1)
- _ 1 000 _ _ _ _ _ - 3 places for 111 to come - 3*8 possible strings
- _ _ 1 000 _ _ _ _ - 2 places for 111 to come - 2*8 - 2 possible strings (excluding 111 at beginning as this would be counted later)
- _ _ _ 1 000 _ _ _ - 1 places for 111 to come - 8 - 2 possible (excluding 1111000 and 01111000 as these would be counted later)
So, totally $80+32+24+14+6 = 156$ possible strings and by symmetry we should have $160$ strings for the case where 111 comes before 000. Thus, we should subtract $2*156=312$ from our earlier count of 1040, which gives $1040 - 312 = 728.$
But we are not yet done, there might be strings with "000" or "111" repeated and those we have subtracted twice. Lets consider "000" being repeated.
000 000 _ _ _ _ - not happening in our count
000 1 000 111 - happens
000 1111 000 - happens
0000111000 - happens
Similarly 111 0 111 000, 111 0000 111 and 1111000111 also happens. So, we must add 2*3=6 to our answer which gives
$$728+6=734$$