There are 4 spots to start the word, and then there are \(3!\) ways to arrange the other letters in the remaining three spots. Thus the number of words avoiding the sub-word “bad” in consecutive letters is \(6! - 4\cdot 3!\text{.}\)
If we now need to avoid words that put “b” before “a” before “d”, we must choose which spots those letters go (in that order) and then arrange the remaining three letters. Thus, \(6! - {6 \choose 3}3!\) words.