Cross Twice or More

We will proceed with complementary counting: count the number of ways to get to the end with zero or one crossing, and subtract it from the total number of paths possible.

First, the total number is $\dbinom{20}{10}=184756$ .

Now, let's count the number of paths that do not cross. But we recognize this as a famous problem of "how many paths without going above the diagonal" with the answer simply $C_{10}$ , where $C_n$ is the $n$ th Catalan number. But note that we can either be always below the diagonal or always above it so there are actually $2C_{10}$ ways.

Now, let's count the number of paths that cross exactly once. If the point of crossing is at $(k,k)$ , then we have divided the problem into two smaller problems of "how many paths without going above the diagonal". Thus, for this case there are $C_k\cdot C_{10-k}$ ways. But note that again, we can start out above the diagonal or below the diagonal, so there are $2C_k\cdot C_{10-k}$ ways for each $1\le k\le 9$ .

Thus, the number of ways for the path to not cross or cross once is $2C_{10}+\sum_{k=1}^92C_k\cdot C_{10-k}=2\sum_{k=1}^{10}C_k\cdot C_{10-k}$

How do we calculate this? We notice that $\displaystyle C_{11}=\sum_{k=0}^{10}C_k\cdot C_{10-k}$ by the recurrence relation of $C_n$ so $2\sum_{k=1}^{10}C_k\cdot C_{10-k}=2(C_{11}-C_0\cdot C_{10}) = 2\left(\dfrac{1}{12}\dbinom{22}{11}-\dfrac{1}{11}\dbinom{20}{10}\right)=83980$

So our final answer is $184756-83980=\boxed{100776}$