Given the following question, quoted verbatim:
How many permutations of all the letters $ABCDEFGH$ contain the strings $CAB$ and $FAD$?
Would the answer be $0$? Because the way I understand it, permutations of a set of distinct items by definition don't include repeats unless otherwise specified, hence why the permutations of $n$ items is calculated by $n!$ and not $n^n$, and a permutation of that set of letters would have to contain $A$ twice to contain both $CAB$ and $FAD$.
Assuming that the letters in the specified strings needn't be consecutive, it should be clear that the specified letters alone can be permuted in $4$ ways, e.g. $FCABD, CFABD,$ etc.
Insert the remaining $3$ letters one by one. The first one can be inserted in $6$ ways at the uparrows, e.g. $\uparrow F\uparrow C\uparrow A\uparrow B\uparrow D\uparrow$, the next one in $7$ ways, and the final one in $8$ ways
Putting everything together, we get $4\cdot6\cdot7\cdot8$ ways