Hopping to the Right

There are 3 intermediate steps between A and B. For each step, we either land there or do not land there.
Once we know which steps we land on, there is a unique way to get from A to B (basically taking the arc to the next step). This sets up the bijection between ways from A to B and the set of steps that we land on.
Hence, there are $2^ 3$ ways to get from A to B.

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We see that this solution easily generalizes to get $2^n$ ways from A to B (where $n$ is the number of intermediate steps).

Essentially, you can get to any point from any previous point. So, the number of ways to get to any point is given by the sum of the ways to get to any of the previous points.

So, if $n_m =$ The number of ways to get to the $m$ th point.

Then $n_m = \sum_{i<m} n_i$

And clearly $n_1 = 1$ and $n_2 = 1$

Therefore,

• $n_1=1$
• $n_2=1$
• $n_3=2$
• $n_4=4$

$n_5 = \boxed8$