Probability that 2 heads do not come consecutively. 
A fair coin is tossed $10$ times. Then the probability that two heads do not appear consecutively is?

Attempt:
Cases are:
Given condition cannot be met with $10, 9, 8, 7$ or $6$ heads.
1) 5 heads + 5 tails.
First fulfilling the essential condition we get:
$\mathrm{HTHTHTHTH}$
Then we'll be left with 1 tail which can be placed at 6 places.
So number of cases = $6$
2) $4$ heads, $6$ tails
$\mathrm{HTHTHTH}$
Left with $3$ tails.
So all $3$ can be placed together at $5$ places.
Or $3 = 2+1$, in which case we have $5 \times 4 = 20$ cases.
Total number for #2 $= 20 + 5 = 25$
3) 3 heads, 7 tails.
$\mathrm{HTHTH}$
Left with $5$ tails.
All 5 can be placed together, so in that case we have $4$ cases.
Or $5 = 2+3= 1+4$
For $5 = 2+3 = 1+4$, we have $4\times 3$ cases each.
Therefore total number of cases for #3 $= 12+12+ 4 = 28$
4) 2 heads 8 tails.
$\mathrm{HTH}$
$7$ left
$7 = 7+ 0$ , $3$ cases.
$7 = 6+1= 5+2= 3+4$ , $6$ cases each
So number of cases for #4 = $6\times 3 + 3 = 21$
5) 1 head 9 tail, 10 cases
6) 0 heads, 1 case.
So total number of cases = $6+25+28+21+10 +1  = 91$
I tried thrice and got $91$ cases only.
So answer should be $P(E) = \dfrac{91}{2^{10}}$
But that's not the right answer.
Please tell me my mistake.

Another way: Let the number of solutions for $n$ tosses be $a_n$. For $n \ge 2$, the solutions are either T plus a solution for $n - 1$ tosses, or HT plus a solution for $n - 2$ tosses. So $a_n = a_{n-1} + a_{n-2}$. Since $a_1 = 2$ and $a_2 = 3$ we see that $a_n$ is the Fibonacci number $F_{n+2}$, which makes the probability of success $F_{n+2}/2^n$. For $n = 10$ this gives the answer found by Key Flex.
