5 heads or 5 tails in a row on 10 toss : why this answer? An acquaintance of mine came up with this question:

What is the probability of having 5 heads or 5 tails in a row, when tossing a fair coin 10 times.

Using this answer, I came up with this solution
$\dfrac{10 \choose 5}{2^{10}} = \dfrac{252}{1024} = \dfrac{63}{256}$.
Yet when thinking of it, I cannot explain in plain words my solution.
Can anyone help?
Update: 

My initial answer was flawed
It should have been this instead 
$\dfrac{112}{1024}$
 A: $\binom{10}{5}=\frac{10!}{5!5!}$ counts all the anagrams of the word $HHHHHTTTTT$ which has $10$ letters, $5$ $H$s and $5$ $T$s. $(1/2)^{10}$ is the probability that one of these anagrams appears after $10$ tosses.
However in your question it is asked to count the arrangements where the 5 heads or 5 tails are in a row such as $TTHHHHHTTT$ or $HHHHTTTTTH$, but also $HTHHHHHHTH$ (which is not an anagram of $HHHHHTTTTT$). So your answer is not correct.
In order to have at least $5$ heads in a row, we consider the following cases:
1) $HHHHHTXXXX$, $THHHHHTXXX$, $XTHHHHHTXX$, $XXTHHHHHTX$, $XXXTHHHHHT$, $XXXXTHHHHH$ $\implies 2^{4}+4\cdot 2^3+2^4=64$;
2) $HHHHHHTXXX$, $THHHHHHTXX$, $XTHHHHHHTX$, $XXTHHHHHHT$, $XXXTHHHHHH$ $\implies 2^{3}+3\cdot 2^2+2^3=28$;
3) $HHHHHHHTXX$, $THHHHHHHTX$, $XTHHHHHHHT$, $XXTHHHHHHH$ $\implies 2^{2}+2\cdot 2^1+2^2=12$;
4) $HHHHHHHHTX$, $THHHHHHHHT$, $XTHHHHHHHH$ $\implies 2^{1}+2^0+2^1=5$;
5) $HHHHHHHHHT$, $THHHHHHHHH$  $\implies 2^0+2^0=2$;
6) $HHHHHHHHHH$ $\implies 2^0=1$;
and the total number is $64+28+12+5+2+1=112$.
Similarly the total number of ways to have at least 5 tails in a row is $112$. For a row of $5$ heads OR $5$ tails the number of ways is 
$112+112-2=222$  (we subtract $2$ because otherwise $TTTTTHHHHH$ and $HHHHHTTTTT$ are counted twice).
Finally the probability of having $5$ heads or $5$ tails in a row, when tossing a fair coin $10$ times is
$$\frac{222}{2^{10}}=\frac{111}{512}.$$
