We all know that
$$C_n = \frac{1}{n+1} {2n \choose n} $$
But what if I want to calculate the same property of catalan, but with number of zeros is $s$ and number of ones is $t$ when $s\neq t$? As in:
Calculate in how many binary sequences with length of $s+t$ there are exactly $s$ zeros, and in all sub-sequences the number of ones ($t$) is not greater than the number of zeros ($t\leq s$)?
In my opinion, to calculate this we might need to assign a function $F: \text{group of all bad sequences} \longrightarrow \text{group of all binary sequences of s+1 ones and t-1 zeros}$
But I have no idea on how to go on!
This is the generalization (or perhaps the original formulation) of Bertrand's Ballot Problem. The number of such paths is given by the reflection principle and is $$C_{s,\ t} = \frac{s-t+1}{s+1}\binom{s+t}{t}$$ Notice that when $s=t=n$ then we recover the original Catalan numbers $$C_{n,n}=C_n=\frac{1}{n+1}\binom{2n}{n}$$