In how many ways can we add $1$'s and $2$'s to get $11$ (when the order matters)? Examples:
$1+1+1+1+1+1+1+1+1+1+1$,
$2+2+2+2+2+1$, 
$1+2+2+2+2+2$ (order matters)
I tried solving it with permutations but I realized it won't work
 A: Our first rather clumsy solution takes advantage of the fact that $11$ is a quite small number. 
We could have $0$ $2$'s, $1$ way.
We could have $1$ $2$ and $9$ $1$'s, total $10$ numbers. The location of the $2$ can be chosen in $\binom{10}{1}$ ways.
We could have $2$ $2$'a and $7$ $1$'s, total of $9$ entries. The location of the $2$'s can be chosen in $\binom{9}{2}$ ways.
We could have $3$ $2$'s and $5$ $1$'s. There are $\binom{8}{3}$ choices.
For $4$ $2$'s there are $\binom{7}{4}$ choices, and for $5$ there are $\binom{6}{5}$ choices.
Add up. 
Another way: Let $a_n$ be the number of ways to represent $n$ as an ordered sum of $1$'s and/or $2$'s. We have $a_1=1$ and $a_2=2$.
There are two types of sequences of length $n+1$, (i) the ones that end in $1$ and (ii) the ones that end in $2$. There are $a_n$ of Type (i), and $a_{n-1}$ of Type (ii). It follows that
$$a_{n+1}=a_n+a_{n-1}.$$
We can now use the above recurrence to find $a_3$ then $a_4$ and so on. We get the familiar Fibonacci sequence.
A: I think I've seen a similar question somewhere, and I guess finding the coefficient of $x^{11}$ for $(x^1+x^2)^k$ over all $k$ should do the trick.
Reasoning:
Each new term you pick either exponent 1 or 2, and they add up in the exponent thanks to the binomial theorem.
