The given digits are 4, 6, 9, 5, 3, \( x \), and \( y \).
These are distinct digits, and \( x, y \in \{0, 1, 2, 7, 8 \} \) since they must be distinct from the given five digits.
Also \( x \neq y \).
For a number to be divisible by 3, the sum of its digits must be divisible by 3.
Sum of the given five digits = \( 4 + 6 + 9 + 5 + 3 = 27 \).
The sum of all seven digits is \( 27 + x + y \).
For this sum to be divisible by 3, \( x + y \) must be divisible by 3.
The possible pairs of distinct digits \( (x, y) \) from \( \{0, 1, 2, 7, 8 \} \) such that \( x + y \) is divisible by 3 are:
\begin{itemize}
\item \( 0 + 3 \) (3 not available)
\item \( 0 + 6 \) (6 not available)
\item \( 1 + 2 = 3 \) \( \implies (1, 2), (2, 1) \)
\item \( 1 + 5 \) (5 not available)
\item \( 1 + 8 = 9 \) \( \implies (1, 8), (8, 1) \)
\item \( 2 + 4 \) (4 not available)
\item \( 2 + 7 = 9 \) \( \implies (2, 7), (7, 2) \)
\item \( 7 + 8 = 15 \) \( \implies (7, 8), (8, 7) \)
\item \( 0 + ? \) (0, 3, 6, 9 - 3, 6, 9 not available)
\end{itemize}
The pairs \( (x, y) \) from \( \{0, 1, 2, 7, 8 \} \) such that \( x + y \) is divisible by 3 are:
\begin{itemize}
\item \( (0, 3) \) - not possible
\item \( (1, 2) \), sum = 3
\item \( (1, 8) \), sum = 9
\item \( (2, 1) \), sum = 3
\item \( (2, 7) \), sum = 9
\item \( (7, 2) \), sum = 9
\item \( (7, 8) \), sum = 15
\item \( (8, 1) \), sum = 9
\item \( (8, 7) \), sum = 15
\end{itemize}
The ordered pairs \( (x, y) \) are: \( (1, 2), (2, 1), (1, 8), (8, 1), (2, 7), (7, 2), (7, 8), (8, 7) \).
There are 8 such ordered pairs.