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.
 A: Given that a fair coin is tossed $10$ times. So, the possible outcomes are $2^{10}$ 
Now to find the probability that no two heads appear consecutively, then the possible outcomes are 
$$\begin{matrix} Heads & Tails & Total \\ 0 & 10 & \dbinom{11}{0} \\ 1 & 9 & \dbinom{10}{1} \\ 2 & 8 & \dbinom{9}{2} \\ 3 & 7 & \dbinom{8}{3} \\ 4 & 6 & \dbinom{7}{4} \\ 5 & 5 & \dbinom{6}{5} \\  \end{matrix}$$
We shouldn't consider the cases where heads are $>6$, otherwise there will be two heads consecutively. So, we neglect those cases.
Since, all the outcomes are mutually exclusive$$\left(1+\dbinom{10}{1}+\dbinom{9}{2}+\dbinom{8}{3}+\dbinom{7}{4}+\dbinom{6}{5}\right)\left(\dfrac{1}{2^{10}}\right)$$
$$(1+10+36+56+35+6)\left(\dfrac{1}{2^{10}}\right)=\dfrac{144}{2^{10}}$$
A: You are missing the cases where the tails can be split into more than 2 groups.
i.e. 3 tails can be split into 1+1+1, yielding another 10 cases = $ {5 \times 4 \times 3\over 3 \times 2 \times 1} $ etc...
The final answer should be $144\over2^{10}$. There are 28 more cases for 3 heads and 15 more cases for 2 heads.

Key Flex's answer is a more straightforward way to calculate the answer. To think about it more intuitively, instead of breaking tails into differently sized groups to place between the heads, break the heads into individual groups of 1 to slot between the tails.
A: 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.
