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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 $FAD$.

user3776749
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    You’re right that repetitions are not allowed, but you may be incorrectly interpreting contain the strings: it may be that all substrings are intended, not just substrings of consecutive letters, in which case $CFABDEGH$ contains both $CAB$ and $FAD$. – Brian M. Scott Nov 29 '16 at 21:32
  • Hm Interesting. Thanks for the tip, I'll look into that. But to clarify, if the letters do need to be consecutive, the answer to that specific question would be zero? – user3776749 Nov 29 '16 at 21:35
  • It would indeed. – Brian M. Scott Nov 29 '16 at 21:36

1 Answers1

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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

true blue anil
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