All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: solution to part 1; thoughts on part 2 | Comment 7 of 8 |
(In reply to solution to part 1; thoughts on part 2 by Charlie)

Not sure what significance it has, if any, but the only other "pattern" I see in the higher columns of your table is that n(a,s) is evenly divisible by 2(a-s+1).  Unfortunately I don't see a pattern in the rest of the divisors that would lead to a closed-form solution for n(a,s).  For example, for n(a,5):

n(5,5) = 14 = 2*7

n(6,5) = 124 = 4*31

n(7,5) = 582 = 6*97

n(8,5) = 1928 = 8*241

n(9,5) = 5110 = 10*511 = 10*7*73

n(10,5) = 11604 = 12*967

n(11,5) = 23534 = 14*1681 = 14*41*41

n(12,5) = 43792 = 16*2737 = 16*7*17*23

Edited on March 26, 2014, 9:04 am
  Posted by tomarken on 2014-03-26 09:01:02

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (0)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information