Find all ordered pairs $(a,b)$ of positive integers that satisfy $a>b$  and the equation $(a-b)^{ab}=a^bb^a$.
We are given the equation \((a-b)^{ab} = a^b b^a\) and the condition \(a > b\) where \(a\) and \(b\) are positive integers. We need to find all ordered pairs \((a, b)\) that satisfy these conditions.

1. **Case 1: \(v_2(a) > v_2(b)\)**

   - Let \(v_2(a) = k\) and \(v_2(b) = l\) with \(k > l\).
   - Then \(v_2(a - b) = \min\{v_2(a), v_2(b)\} = v_2(b) = l\).
   - Therefore, \(v_2((a - b)^{ab}) = ab \cdot v_2(a - b) = ab \cdot l\).
   - On the other hand, \(v_2(a^b b^a) = v_2(a^b) + v_2(b^a) = b \cdot v_2(a) + a \cdot v_2(b) = b \cdot k + a \cdot l\).
   - Equating the two expressions, we get:
     \[
     ab \cdot l = b \cdot k + a \cdot l
     \]
   - Rearranging, we have:
     \[
     ab \cdot l - a \cdot l = b \cdot k
     \]
     \[
     a \cdot l (b - 1) = b \cdot k
     \]
     \[
     \frac{a}{k} = \frac{b}{b - 1}
     \]
   - Since \(a\) and \(b\) are integers, \(\frac{a}{k}\) must be an integer. Let \(a = 2^m\) and \(b = 2^n\) where \(m > n\). Then:
     \[
     \frac{2^m}{m} \leq \frac{2^n}{n - 1}
     \]
   - It is obvious that \(\frac{2^m}{m} \geq 2\), so \(b\) must be 2. Therefore, \(a = 2v_2(a)\). Since \(a > b = 2\), we have \(a = 4\).

2. **Case 2: \(v_2(a) < v_2(b)\)**

   - Similarly to Case 1, we would get \(a = 2\), which contradicts \(a > b\).

3. **Case 3: \(v_2(a) = v_2(b) = c\)**

   - Let \(a = 2^c \cdot a'\) and \(b = 2^c \cdot b'\) where \(a'\) and \(b'\) are odd.
   - Substituting into the original equation, we get:
     \[
     (2^c(a' - b'))^{ab} = (2^c)^{ab}(a' - b')^{ab} = 2^{abc}(a' - b')^{ab}
     \]
     \[
     a^b b^a = (2^c a')^b (2^c b')^a = 2^{bc} a'^b 2^{ac} b'^a = 2^{a+b} a'^b b'^a
     \]
   - Since \(ab > a + b\) (because \(a > b\) and both are even), this leads to a contradiction.

Thus, the only solution is \(a = 4\) and \(b = 2\).

The final answer is \(\boxed{(4, 2)}\).