How many ways can this hexagon be tiled by 11 rhombuses of unit side length? I came across a question in an exercise booklet for Mathematic Olympiad for primary school students in Australia. The question is shown in the following picture:

I barely have any clue how this sort of questions can be approached. Would anyone please help solve this question and possibly suggest a general solution, if any, to this kind of problems. Thank you very much!
 A: If you put a single rhombus vertically, the rest of the rhombuses on either side have to be placed horizontally, otherwise there will be some "lonely" triangles that don't get covered. (Try putting the vertical rhombus on the far rightmost side first, and then fill in from there, to convince yourself of this). Also, there has to be at least one vertical rhombus in the tiling; placing all the rhombuses horizontally leaves "lonely" triangles again. So the number of possible tilings is equal to the number of places you can initially put the vertical rhombus, which in this case is 6.
A: Let's denote our hexagon-like figures as $H(n)$, where $n$ is the horizontal length of it. In our case $n=6$.
Let's also denote number of tiling of such rhombus by $T(n)$.
See, that for $n>1$ we can start tiling in two ways:

*

*By placing single rhombus vertically:


This option gives us only one way of evaluation - adding rhombuses 'horizontally':



*By placing two rhombuses 'horizontally':


As we can see, by this action we've reached the figure $H(n-1)$
For $n=1$ it's obviously only one way of tiling.
Therefore we can describe the number of tilings using the following formula:
$$T(n) = \begin{cases}1, & n=1\\ T(n-1)+1 , & n>1\end{cases}$$
This is simply equivalent to the neutral function:
$$T(n)=n$$
Therefore in our case ($n=6$) the number of possible tilings is $6$
