I am trying to understand a proof to come up with the catalan numbers presented in the book "A course in combinatorics" by van Lint and Wilson. The authors say that by reflecting the part of the path between A and the first meeting with the X-axis with respect to the X-axis, we find a path from the reflected point A' to B. THey claim to estabish a one-to-one correspondence between paths from A' to B and paths from A to B that meet or cross the X-axis.
Then they say that if follows that if $A=(0,k)$ and $B=(n,m)$, then there are ${n \choose l_1}$ paths from A to B that cross or meet the X-axis, where $2l_1 = n-k-m$. Since there are ${n \choose l_2}$ paths from $A$ to $B$, where $2l_2 = n-m+k$, we find ${n \choose l_2} - {n \choose l_1}$ paths from $A$ to $B$ that do not meet the X-axis. What I am confused about is how they determined the $l_1$ and $l_2$, I don't really understand how they came up with those formulas. Could anyone explain this to me?
To go from $A'$ to $B$ you take $n$ steps, each of them either upwards or downwards. Suppose $l_1$ of them go downwards; then the remaining $n-l_1$ steps go upwards. The net upwards movement is thus $(n-l_1)-l_1 = n-2l_1$, which must equal the vertical distance from $A'=(0,-k)$ to $B=(n,m)$, i.e., $m-(-k)=m+k$. So $$ n-2l_1 = m+k , $$ which gives the formula for $l_1$. The number of paths is then $\binom{n}{l_1}$, since this is the number of ways you can choose which $l_1$ of the $n$ steps that go downwards.
Similarly for $l_2$, if you use $A=(0,k)$ instead of $A'=(0,-k)$ (i.e., just replace $k$ by $-k$ in the previous formulas).