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My work:

$3$ (as there are $3$ choices for first digit) x $2^9$ (only $2$ digits as choices to avoid consecutive digits from then on) x $10$ (permutations where we move the 3 around) = 15360 sequences

But other people are saying that the answer is just $3$ x $2^9$. Can't we move the $3$ around to get $10$ different permutations?

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    Other people are right. Think of a case with only (0,1) to choose. There are clearly only 2 sequences and not 20. – cr001 Oct 07 '20 at 05:48
  • Forgive me for asking, so something like 2 x 3 x 2^8 would be strange because we know (logically) that a sequence can start with 3 numbers (0, 1, 2) so it's illogical to think of it this way where we have two choices to start with correct? –  Oct 07 '20 at 05:49
  • Oh wait now that I think about it, it makes sense as if we start with 2 then we continue with 2 and would never have a choice of 3 numbers to pick from via the condition stated in the question never mind thank you! –  Oct 07 '20 at 05:50
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    @TotomaLover2312 but all those cases are covered in $3 \times 2^9$ – Math Lover Oct 07 '20 at 06:06
  • Yeah I see that now. Thanks! –  Oct 07 '20 at 06:06

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