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Re: Permutation/Combination Problem
Posted:
Feb 7, 2013 1:28 AM


On Feb 6, 6:29 pm, 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)(N2)*(3^(N3)) 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.
The inandout principle leads directly to the following formula:
sum, from k = 0 to floor(N/3), of (1)^k * C(N2k,k) * 3^(N3k).
This is sequence A076264, "Number of ternary (0,1,2) sequences without a consecutive '012'", at oeis.org, the online encyclopedia of integer sequences:
http://oeis.org/search?q=1%2C3%2C9%2C26%2C75%2C216%2C622&sort=&language=english&go=Search



