Note that if you were allowed to reuse the digits then the number of possible outcomes would be
\begin{align*}
\frac{13!}{4!9!} & = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} \\
& = 715
\end{align*}
which once again is more since numbers are allowed to repeat.