How many bit strings of length 8 contain either three consecutive 0's or four consecutive 1's ?
MY APPROACH :-
Initially, for 3 consecutive 0's:
000_ _ _ _ _ =>2^5 = 32 WAYS
1000_ _ _ _ =>2^4 = 16 WAYS
_1000_ _ _ =>2^4 = 16 WAYS
_ _1000_ _ =>2^4 = 16 WAYS
_ _ _1000_ =>2^4 = 16 WAYS
_ _ _ _1000 =>2^4 = 16 WAYS
So totally 112 ways
Similarly, for 4 consecutive 1's
1111 _ _ _ _ =>2^4 = 16 WAYS
01111 _ _ _ =>2^3 = 8 WAYS
_01111_ _ =>2^3 = 8 WAYS
_ _01111_ =>2^3 = 8 WAYS
_ _ _01111 =>2^3 = 8 WAYS
So totally 64 ways
Intersections possible:-
0001111X =>2*3! = 12 WAYS
so, total ways = 112 + 48 - 12 = 148 ways
But answer is given as 147 ways.
Where am I wrong?