We have a $5$ step tree and can only move $1$ step up or $1$ step down. If we wanted all the paths that end in the top yellow number, how many paths would that be? Example paths can be $uuddd$, $ddudu$, etc, where $u$ is up and $d$ is down. Would be nice to have a general formula.
Manually finding the paths, I get $10$ and I realised all the paths have $2$ ups and $3$ downs. So this can be $5C3$ or $5C2$ which both give $10$. This then can be extended to saying if $i$ is the position of the number, so the top yellow number is position $i=4$, we can say in general the number of paths to position $i$ in the final branch is $5$ choose $(5-(i-1))$. I made this up so don't know if it is legit.
Another way is to look at the options we have going backwards (going forwards works too but going back seems easier). When we are on $9.75$, we have $2$ options to go back to the previous branch. Then each of those figures ($0.00$ and $20.90$) have $2$ options each, so $4$ in total. Then on and on which gives, $2+4+5+4+2=17$ which contradicts my first method.
Please help! Thanks in advance
Your reasoning is exactly correct.
If you want to get to position $i$ in column $k$, you can divide that paths that do so into
Those that finish with a "d", and come from position $i$ in column $k-1$, or
Those that finish with a "u", and come from position $i-1$ in column $k-1$.
(I'm counting "position" up from the bottom, which is position $0$, and columns from the left, where the leftmost column (with one node) is column $0$.)
If we let $P(k, i)$ be the number of paths from the start to position $i$ in column $k$, then the division above says that $$ P(k, i) = P(k-1, i) + P(k-1, i-1) $$ which is exactly the recurrence formula for the combinatorial coefficients. Because "along the edge" (i.e., for $i = 0$ or $i = k$ in each column $k$) the values are all "1"s (representing "all u" or "all d" paths), the boundary conditions for the recurrences are the same as well. Hence you can conclude that $$ P(k, i) = {{k} \choose{i}}. $$