Counting and Probability Problem The question I am having trouble with is how many permutations of six letters{A,B,C,D,E,F} are there that contains neither "BAD" nor "DEF" patterns.
My plan for solving this would be to find the total number of permutations that can be achieved and subtracting the occurrences of "BAD" and "DEF" patterns from that, but I am not sure how to achieve it. Any help is appreciated.
 A: Think of BAD as one letter. You have three more. So there are 4! arrangements for these 4 letters. Similarly for DEF.
However there is an overcount. We are counting BADEF twice because we count it in each of our 4! So we must subtract 2 (the cases of C BADEF and BADEF C).
So now we can compute the total number which is 6! and subtract that bad sequences which is 4! + 4! -2
A: The number of permutations containing BAD in any three positions in order is $\binom{6}{3}3!$, and similarly the number of permutations containing DEF is $\binom{6}{3}3!$ (pick three spots to put the pattern then arrange the rest of the letters around it). The number of permutations that contain both BAD and DEF is $\binom{6}{3,3}=\binom{6}{3}$. Thus the number of permutations that contain either BAD or DEF in any three positions in order is
$$2\cdot \binom{6}{3}3!-\binom{6}{3}=11\binom{6}{3}$$
A: Use the Law of Complements and the Principle of Inclusion and Exclusion.
Count the total permutations, subtract the counts of those with BAD or DEF, add back the permutations with both BAD and DEF.
$$\begin{align}
|T\setminus (B\cup D)| & = |T| -|B|-|D|+|B\cap D|
\\ & = 6! - 4! - 4! + 2!
\\ & = 674
\end{align}$$
Note: To count of permutations containing DEF (or BAD) permutations of the pattern and 3 other symbols. Ie: {BAD,C,E,F}  and {DEF,A,B,C}.  Hence $4!$ each for $|B|$ and $|D|$
NOTE 2: To count the permutations that contain both, you must count the permutations of {BADEF,C}.  Hence $2!$ for $|B\cap D|$
