How many barcodes are there? A barcode is made of white and black lines. A barcode always begins and ends width a black line. Each line has is of thickness 1 or 2, and the whole barcode is of thickness 12. 
How many different barcodes are there (we read a barcode from left to right).
I know that I'm required to show some effort, yet I really don't have a clue about this problem. Can you give me any hints? 
 A: Lets create an additional variable $k > 6$ denoting the number of bars. It is necessarily an odd number, as black and white bars must alternate. It is larger that $6$ because of your condition on thickness of bars to be $\le 2$. 
Having this number, we can calculate the total number of barcodes with $k$ bars. 
Let each of this bars has thickness $1$. Then we have $12- k$ additional places for an increasing of thickness of any of the existing bars. So, we can choose $12 - k$ bars and enlarge them. We can do this by $C_{k}^{12-k}$ variants, so the answer will be
$$\sum_{k=7,9,11}C_{k}^{12-k}$$
A: Here  is variant  with  some combinatorial algebra.


*

*We encode  the  thickness  $1$ and  $2$ of the black    lines   as  $x^1+x^2$, the exponent  indicating the thickness, the $+$ indicating the alternatives. 

*We  do the same for  the  white  lines and get $y^1+y^2$.

The length  of a  barcode starting  and ending with black  lines can therefore  be encoded as
  \begin{align*}
(x+x^2)(y+y^2)(x+x^2)\cdots (x+x^2)\tag{1}
\end{align*}
Note that beginning and end of this expression is $x+x^2$ since we start and end a barcode with black lines.

A barcode with $k$ white lines has necessarily $k+1$ black lines and is encoded as expression
\begin{align*}
(x+x^2)^{k+1}(y+y^2)^{k}\tag{2}
\end{align*}

We are not interested in differentiating black and white lines. So, we count them all using one variable $z$ and obtain instead of (2)
  \begin{align*}
(z+z^2)^{2k+1}\tag{3}
\end{align*}
Since a barcode may consist of one or more black lines and we are interested in barcodes of length $12$ we add up all expressions of the form (3) and select the coefficient of $z^{12}$.

We use the notation $[z^n]$ to denote the coefficient of $z^n$ in a series.

We obtain this way
  \begin{align*}
[z^{12}]\sum_{k\geq 0}(z+z^2)^{2k+1}
&=[z^{12}]\sum_{k\geq 0}z^{2k+1}(1+z)^{2k+1}\tag{4}\\
&=\sum_{k=0}^{5}[z^{11-2k}](1+z)^{2k+1}\tag{5}\\
&=\sum_{k=3}^5\binom{2k+1}{11-2k}\tag{6}\\
&=\binom{7}{5}+\binom{9}{3}+\binom{11}{1}\\
&=116
\end{align*}

Comment: 


*

*In (4) we factor out $z^{2k+1}$.

*In (5) we use the rule $[z^p]z^qA(z)=[z^{p-q}]A(z)$ and  restrict the sum with upper limit $k=5$, since the exponent $11-2k$ is non-negative.

*In (6) we select  the coefficient of $z^{2k-1}$ and start with index $k=3$ since the other binomial coefficients with $k<3$ are zero.
A: Let $1$ denote a black line & $0$ denote a white line. A barcode is then a sequence of $1$'s & $0$'s that starts & ends with $1$ and avoids $000$ &$111$. So your problem is to find the number of binary words of length $12$ that start and end with $1$ and avoid $000$ & $111$.
Let $a_n$ denote the number of binary words that start with $1$ and end $11$ (& satisfy the $000$, $111$ condition). Let $A(x)$ denote the generating function for $a_n$. Define B(x) similarly to $A$ to end $01$.  Define C(x) similarly to $A$ to end $00$.Define D(x) similarly to $A$ to end $10$. These functions are related by the following recurrence relations.
\begin{eqnarray*}
A=x^2+xB \\
B=x(C+D) \\
C=xD \\
D=x^2+x(A+B)
\end{eqnarray*}
Now solve these and find the coefficients $x^{12}$ in $A$ & $B$. Add these value.
A: Denote by $b_i$ the number of bars (black or white) of width $i\in\{1,2\}$. Then $b_1+2b_2=12$, hence $b_1$ is even. Since the total number of bars $b_1+b_2$ is odd it follows that $b_2$ is odd. This leaves the cases
$$(b_1,b_2)\in\bigl\{(10,1),(6,3),(2,5)\bigr\}\ .$$
The total number of admissible arrangements then comes to
$${11\choose 1}+{9\choose 3}+{7\choose 2}=116\ .$$
