How many non-intersecting pairs  of  paths we have from (0,0) to (n,n) so that path can move two ways:top or right?
1. **Understanding the Problem:**
   We need to find the number of non-intersecting pairs of paths from \((0,0)\) to \((n,n)\) where each path can move either up (\(\uparrow\)) or right (\(\rightarrow\)).

2. **Total Number of Paths:**
   Each path from \((0,0)\) to \((n,n)\) consists of \(n\) upward moves and \(n\) rightward moves. The total number of such paths is given by the binomial coefficient:
   \[
   \binom{2n}{n}
   \]
   This is because we need to choose \(n\) positions out of \(2n\) for the upward moves (or equivalently, the rightward moves).

3. **Counting Intersecting Pairs:**
   If two paths intersect, one path starts with \(\uparrow\) and ends with \(\rightarrow\), while the other path starts with \(\rightarrow\) and ends with \(\uparrow\). Consider two paths:
   - Path 1: \((1,0) \rightsquigarrow (n,n-1)\)
   - Path 2: \((0,1) \rightsquigarrow (n-1,n)\)

   The total number of such intersecting pairs of paths is:
   \[
   \binom{2n-2}{n-1}^2
   \]
   This is because each path from \((1,0)\) to \((n,n-1)\) and from \((0,1)\) to \((n-1,n)\) consists of \(n-1\) upward moves and \(n-1\) rightward moves.

4. **Counting Non-Intersecting Pairs:**
   To count the number of non-intersecting pairs, we need to subtract the number of intersecting pairs from the total number of pairs. The number of intersecting pairs can be found by considering a bijection that switches the path until its first intersection. The number of such intersecting pairs is the same as the number of path pairs:
   - Path 1: \((1,0) \rightsquigarrow (n-1,n)\)
   - Path 2: \((0,1) \rightsquigarrow (n,n-1)\)

   The number of such intersecting pairs is:
   \[
   \binom{2n-2}{n-2}^2
   \]

5. **Final Calculation:**
   Therefore, the number of non-intersecting pairs of paths is:
   \[
   \boxed{\binom{2n-2}{n-1}^2 - \binom{2n-2}{n-2}^2}
   \]