Upstairs

$\displaystyle \sum _{ k=0 }^{ 10 }{ \left( \begin{matrix} 20-k \\ k \end{matrix} \right) } =10946$

Consider the following combinations, and count permutations of each

20 single steps 0 double steps
18 single steps 1 double step
16 single steps 2 double steps
etc.

Let $a_{n}$ be the number of ways of climbing a staircase of $n$ steps, 1 or 2 steps at a time. Clearly $a_{1} = 1$ and $a_{2} = 2$ . Now with $a_{n}$ for $n \gt 2$ , we can take any sequence covering $n - 1$ steps and then rise 1 step more, or we can take any sequence covering $n - 2$ steps and rise 2 steps in one go. With these two scenarios being mutually exclusive, we see that for $n \gt 2$ , $a_{n} = a_{n - 1} + a_{n - 2}$ , which we recognize as a Fibonacci formulation. So working our way up the sequence $1,2,3,5,8,13, .....$ until the 20th step we obtain a desired value of $\boxed{10946}$ .

Let's just start with smaller stairs:

**1 step= 1 possibilty,

2 steps= 2 possibilities,

3 steps= 3 possibilities,

4 steps= 5 possibilities **
Then, we notice that the number of possibilities to climp 5 steps can be given by: (1) If we walk up 1 step at the beginning, there are 4 steps left that we can climb in 5 ways.

* (2) * if we walk up 2 steps at the beginning, there are 3 steps left that we can climb in 3 ways. So the number of possibilities to climb 5 steps= (1) + (2)= possibilities for 4 steps+ possibilities for 3 steps=8 possibilities
The generalization is: $Un=U(n-1)+U(n-2), Uo=1, U1=1$

(With the help of your calculator or by repeating it 20 times) The result is $U20=U19+U18=6765+4181=10946$