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 Words count (Posted on 2014-03-25)
(D3)How many subsets of (a,b,c, … x,y,z) contain no letters neighboring each other in the English ABC (26 letters)?

(D4)How many "words" (i.e, ordered subsets from (a,b,c, … x,y,z)) contain no neighboring letters that neighbor each other in the English ABC (26 letters)?

 No Solution Yet Submitted by Ady TZIDON No Rating

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 Solution | Comment 1 of 8
The first part's pretty simple.  If you consider shorter alphabets:
A 1 letter alphabet has 2: {}, a
A 2 letter alphabet has 3: {}, a, b
A 3 letter alphabet has 5: the 2 from a 1 letter alphabet with c added to it plus the ones from a 2 letter alphabet.
This recurrence sets up the Fibonacci sequence with an of 2 offset.
A 26 letter alphabet has F(28)=317811

Part 2 I can see the recursive structure but it's more complicated.  Though very computable, I stopped at 9.  Basically you have to multiply the sequence length by the factorial of its length.  I'll post my reasoning later but the sequence ends up as https://oeis.org/A072374 and the term for a 26 letter alphabet is 571177352091

 Posted by Jer on 2014-03-25 13:09:33
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