Double counting within sequences and arrangements Say we are forming sequences of numbers in the interval [0-9], lengths of (2x) where x is a value from [1,5].  So we form sequences of 2, 4, 6, 8, and 10 numbers.
A successful sequence is a sequence where at least half of the elements, (at least x of the elements), are the same number consecutively.  What is the total of 2, 4, 6, 8, and 10 number sequences that are successful?
Example: 2234 is successful.  3131 is not. 12 is successful. 2233 is successful. 2223 is successful.
Solution attempt:  My original plan was to choose 1 of the 10 numbers to have x times in a row, so 10 choices.
My second step was to choose the starting location of the x-length piece, and the x-length piece can be started at either the first position, all the way up to the x+1 position in the 2x length sequence.  So there are x+1 choices for the starting point of the x-length sequence.
My third step was to choose 1 of 10 letters for each of the remaining x slots, but now I realize I am double counting.
For example, the sequence 1122 could be obtained by choosing 11 as our x length sequence, placing it in the first slot, and happening to choose 2s to fill the remaining slots.
1122 could also be obtained by choosing 2 for the x length sequence, placing it in the third (x+1) slot, and happening to choose 1s to fill the rest.
I am assuming this is some sort of summation for values of x 1,2,3,4,5 or 2x, 2,4,6,8,10.  However, I am unsure of how to account for double counting in a way that is applicable to the general pattern of the scenario.  This isn't exactly something I feel should be solved with brute force.  Any help would be greatly appreciated.
 A: As you ask for a sequence of length x, there are not many ways to have two sequences of distinct numbers in a sequence of length 2x. Precisely, you can chose 10 numbers for the first sequence, and 9 for the second, so 90 in total.
What is more difficult are longer sequences of the same number, for instance 1112 is counted 2 times, and 1111 even 3 times.
To avoid double counting these sequences, count the number of sequences of length x, and subtract the number of sequences of length x+1. A sequence of length x+k is now counted exactly once: you add it k+1 times when you count the sequences of length x, and subtract it k times when you subtract the sequences of length x+k.
To have the answer you want, you still have to count the possibilities of x and x+1 sequences for every x, subtract the possibilities of two sequences of different numbers and add all together, but I assume this is not considered as brute force.
A: Consider a valid string containing $x$-tuple of twos, and an extra $n\ (0\leq n\leq x)$ number of twos. Thus you end up with exactly $(x-n)$ non-two symbols(of which there are $9^{x-n}$ possibilities) in the string. The $x$-tuple could be anywhere in the $(x-n+1)$ positions between/beyond these symbols.
The only positions left to be enumerated now are for the extra $n$ twos; these can go into any combination of $(x-n+2)$ places between the objects we have so far. Take a look at the partition adjacent to the $x$-tuple on the left and the one on the right. It should not matter if we assign $2$ twos to the one on the left and a single two to the one on the right, the end result is that you end up with a $(x+3)$-tuple of twos. (**Old explanation: This means we are overcounting by a factor of two. So remember to divide the above binomial coefficient by $2$. **Not true when n=0; also not true if both the partitions are assigned zero.) 
So the two partitions may be combined into one. Note that multiple twos may occupy a single position here, so we are looking for the number of weak compositions of $n$ twos into $(x-n+1)$ partitions, which is:
$$
\binom{n+(x-n+1)-1}{n} = \binom{x}{n}
$$
Now put everything together:
$$
10\sum_{n=0}^{x} 9^{x-n}(x-n+1) \binom{x}{n}
$$
This still overcounts by exactly 90. (verified for each of $x=1,2,3,4$)
If you do a few examples by hand you'll notice that it double counts every case where a sequence consists of two parts: an $x$-tuple of one symbol, and an $x$-tuple of another symbol. For example, $x=2, n=0$: The sequence 2233 will be counted twice; once while counting sequences with $x$-tuple of $2$'s and again while counting $x$-tuple of $3$'s.
Accounting for every such permutation, we have: 
$$
10\sum_{n=0}^{x}\left\{ 9^{x-n}(x-n+1) \binom{x}{n} \right\} - \left(10 \right)_2
$$
