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Is it true that given a series of digits there always exists an $n$ where $n!$ contains that series of digits?

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    $\begingroup$ Just to clarify: by "series of digits" do you mean a finite sequence of digits $(a_1,\ldots,a_m)$? $\endgroup$ – parsiad Nov 6 '16 at 15:51
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    $\begingroup$ Is there any reason you have to suspect that this would be true? One can ask the exact same question about any sequence. The answer is usually either no, or we don't know. $\endgroup$ – Mathmo123 Nov 6 '16 at 15:51
  • $\begingroup$ Also, do you mean contain exactly (as a substring, say) or do you just want your digits to appear in that order in $n!$ but might be non-consecutive (like $(4,3,2)$ appearing in the number $40320=8!$)? I assume the former. $\endgroup$ – Angelo Rendina Nov 6 '16 at 15:56
  • $\begingroup$ @parsiad yes, finite sequence, sorry, my English is not always correct. And yes, substring like. $\endgroup$ – chx Nov 6 '16 at 16:00
  • $\begingroup$ @chx: I think Mathmo123's question is also interesting; do you have a reason for suspecting this to be true? (Perhaps you have some insight you can share with potential answerers). $\endgroup$ – parsiad Nov 6 '16 at 16:02

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