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Topic: Permutation/Combination Problem
Replies: 3   Last Post: Feb 7, 2013 2:08 AM

 Messages: [ Previous | Next ]
 gnasher729 Posts: 419 Registered: 10/7/06
Re: Permutation/Combination Problem
Posted: Feb 6, 2013 9:44 PM

On Feb 7, 12:29 am, willmann...@gmail.com wrote:
> Sorry for the vulgarity:  I have a 3 letter keyboard with the letters E, S, & X.
> I need a formula that can tell me how many words of length N that I can make without the word SEX contained in it. The letters can be repeated so if it is of length 4 then SSXX would be ok or SSSS etc.  So my initial approach was to do (3^N) to find the total number of permutations(I think it is a permutation please correct me if it is a combinations) I can come up with and then subtract all those permutations that contain an instance or instances of the words SEX.  Any ideas on a mathematical formula for this?
>
> For anything of length 1 the answer is 3; for length 2 it is 9; length 3,4,and 5 this formula works (3^N)-(N-2)*(3^(N-3)) but anything higher than that does not and I think it has something to do with if it is of length 6 or 7 or 8 the word SEX can appear twice and if it is between length 9 and 11 it can appear 3 times and so on.  Any help is much appreciated.  I am looking for a formula that always works.

I'd start like this:

f (N) = number of sequences not containing S, E, X.
g (N) = number of those sequences ending in S
h (N) = number of those sequences ending in S, E.

g (N + 1) = f (N)
h (N + 1) = g (N)
f (N + 1) = 3 f (N) - h (N)

There is a reasonably simple method to solve that kind of recursions.
The vector f, g, h (N + 1) can be found my multiplying f, g, h (N) by
some constant matrix, so find the eighenvalues of the matrix and
things look good.

Date Subject Author
2/6/13 gnasher729
2/7/13 Butch Malahide
2/7/13 bacle