Given a prime number $p{}$ and integers $x{}$ and $y$, find the remainder of the sum $x^0y^{p-1}+x^1y^{p-2}+\ldots+x^{p-2}y^1+x^{p-1}y^0$ upon division by $p{}$.
To solve the problem, we need to find the remainder of the sum \( S = x^0y^{p-1} + x^1y^{p-2} + \ldots + x^{p-2}y^1 + x^{p-1}y^0 \) upon division by \( p \), where \( p \) is a prime number.

1. **Expression of the Sum**:
   The given sum can be written as:
   \[
   S = \sum_{k=0}^{p-1} x^k y^{p-1-k}
   \]

2. **Using Fermat's Little Theorem**:
   Fermat's Little Theorem states that for any integer \( a \) and a prime \( p \):
   \[
   a^p \equiv a \pmod{p}
   \]
   This implies:
   \[
   x^p \equiv x \pmod{p} \quad \text{and} \quad y^p \equiv y \pmod{p}
   \]

3. **Expression for \( x^p - y^p \)**:
   Consider the polynomial \( x^p - y^p \). Using Fermat's Little Theorem, we have:
   \[
   x^p - y^p \equiv x - y \pmod{p}
   \]

4. **Factoring \( x^p - y^p \)**:
   The polynomial \( x^p - y^p \) can be factored as:
   \[
   x^p - y^p = (x - y)(x^{p-1} + x^{p-2}y + \ldots + xy^{p-2} + y^{p-1})
   \]
   Notice that the sum inside the parentheses is exactly \( S \):
   \[
   x^p - y^p = (x - y)S
   \]

5. **Modulo \( p \) Analysis**:
   Since \( x^p \equiv x \pmod{p} \) and \( y^p \equiv y \pmod{p} \), we have:
   \[
   x - y \equiv x^p - y^p \pmod{p}
   \]
   Therefore:
   \[
   x - y \equiv (x - y)S \pmod{p}
   \]

6. **Case Analysis**:
   - If \( x \equiv y \pmod{p} \), then \( x - y \equiv 0 \pmod{p} \). This implies:
     \[
     0 \equiv 0 \cdot S \pmod{p}
     \]
     which is always true. Hence, \( S \equiv 0 \pmod{p} \).

   - If \( x \not\equiv y \pmod{p} \), then \( x - y \not\equiv 0 \pmod{p} \). Since \( p \) is prime and \( \gcd(x - y, p) = 1 \), \( x - y \) has a multiplicative inverse modulo \( p \). Let \( (x - y)^{-1} \) be the inverse of \( x - y \) modulo \( p \). Multiplying both sides of the congruence \( x - y \equiv (x - y)S \pmod{p} \) by \( (x - y)^{-1} \), we get:
     \[
     1 \equiv S \pmod{p}
     \]

Conclusion:
- If \( x \equiv y \pmod{p} \), then \( S \equiv 0 \pmod{p} \).
- If \( x \not\equiv y \pmod{p} \), then \( S \equiv 1 \pmod{p} \).

The final answer is \( \boxed{ S \equiv 0 \pmod{p} } \) if \( x \equiv y \pmod{p} \) and \( S \equiv 1 \pmod{p} \) if \( x \not\equiv y \pmod{p} \).