Let $ m, n \geq 1$ be two coprime integers and let also $ s$ an arbitrary integer. Determine the number of subsets $ A$ of  $ \{1, 2, ..., m \plus{} n \minus{} 1\}$ such that $ |A| \equal{} m$ and $ \sum_{x \in A} x \equiv s \pmod{n}$.
1. **Define the set and the problem:**
   Let \( m \) and \( n \) be two coprime integers with \( m, n \geq 1 \). We need to determine the number of subsets \( A \) of \( \{1, 2, \ldots, m+n-1\} \) such that \( |A| = m \) and \( \sum_{x \in A} x \equiv s \pmod{n} \).

2. **Define \( A_s \):**
   For an integer \( s \), denote \( A_s \) as the set of \( m \)-tuples \((X_1, X_2, \ldots, X_m)\) where \( 1 \leq X_1 < X_2 < \cdots < X_m \leq m+n-1 \) and \( X_1 + X_2 + \cdots + X_m \equiv s \pmod{n} \).

3. **Construct a bijection:**
   We construct a bijection between \( A_s \) and \( A_{s+m} \):
   - For \((X_1, X_2, \ldots, X_m) \in A_s\), define \((Y_1, Y_2, \ldots, Y_m)\) as follows:
     - If \( X_m < m+n-1 \), let \( (Y_1, Y_2, \ldots, Y_m) = (X_1+1, X_2+1, \ldots, X_m+1) \).
     - If \( X_m = m+n-1 \) and \( m-t \) is the greatest positive integer \( t \) such that \( X_{m-t} < m-t+n-1 \), let \( (Y_1, Y_2, \ldots, Y_m) = (1, 2, \ldots, t, X_{t+1}+t+1, \ldots, X_m+t+1) \).
     - If \( X_m = m+n-1 \) and there doesn't exist a positive integer \( t \) such that \( X_t < t+n-1 \), let \( (Y_1, Y_2, \ldots, Y_m) = (1, 2, \ldots, m) \).

4. **Verify the bijection:**
   It is trivial that \((Y_1, Y_2, \ldots, Y_m) \in A_{s+m}\) and the map from \((X_1, X_2, \ldots, X_m)\) to \((Y_1, Y_2, \ldots, Y_m)\) is a bijection.

5. **Equivalence of sets:**
   Since \((m, n) = 1\), we get that all \( A_s \) have the same number of elements. This is because the bijection shows that the number of elements in \( A_s \) is the same as in \( A_{s+m} \), and since \( m \) and \( n \) are coprime, we can cover all residues modulo \( n \) by adding multiples of \( m \).

6. **Counting the subsets:**
   The total number of \( m \)-element subsets of \(\{1, 2, \ldots, m+n-1\}\) is \(\binom{m+n-1}{m}\). Since there are \( n \) possible values for \( \sum_{x \in A} x \pmod{n} \), each \( A_s \) must have \(\frac{1}{n} \binom{m+n-1}{m}\) elements.

The final answer is \(\boxed{\frac{1}{n} \binom{m+n-1}{m}}\).