Please correct me, if any of my assumptions are wrong.
Suppose I want to find the number of words (real or not) with a length of 10 letters created by our 26-letter alphabet. This would yield $$26^{10}$$ possible combinations.
If I only wanted to use every letter just once the calculation would be $$\frac{26!}{16!}$$ possible combinations. The same result is found with $$26*25*24*23*22*21*20*19*18*17.$$
If repetitions of letters were allowed but no more than two equal letters should be present in a row, I would calculate $$26*25^{9}.$$
How can I calculate the possible number of words constructed from a 26-letter alphabet with length 10, where no triple letters should be present? So, $ABCDEFGHIJ$ would naturally be allowed, as would be $AABBAABBAA$, while $ABCDEEEFGH$ would exceed the maximum number of repeats of a single letter ($E$ in this case).
Based on the previous approaches I would assume $$26*25*24^{8}$$ to be the correct calculation. However, I'm not sure whether it is indeed correct or how to go about finding out, if it is. If I scale it down to shorter words or less letters, it appears to yield lower numbers than expected.
This question is, as I see it, independent of vowels or consonants.
Suppose there are $A(n)$ with the last two letters the same, and $B(n)$ with the last two letters different. In both cases, three-in-a-row is not allowed.
Then $A(n+1)=B(n)$ and $B(n+1)=25(A(n)+B(n))$. So $$B(n+2)=25B(n+1)+25B(n)$$
Look up finite difference equations. The trick is to solve the quadratic equation $$x^2=25x+25$$
then the general solution for $B(n)$ is $$B(n)=Cx_1^n+Dx_2^n$$
You can check that this formula obeys the recursion above. Lastly, calculate $B(1)$ and $B(2)$, then you can work out what $C$ and $D$ must be.