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Words count (Posted on 2014-03-25) Difficulty: 4 of 5
(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    
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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 and the term for a 26 letter alphabet is 571177352091

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