Answer: E

Solution:

The year contributes the digits $2,0,2,3,$ so $2$ appears twice while $0$ and $3$ each appear once. For every digit to end up with an even count, the four digits of the month and day must supply an odd number of $0$'s, an odd number of $3$'s, and an even number of every other digit.

With only four digits available, the month-day string must use exactly one $0,$ one $3,$ and a repeated pair of some digit. Checking valid months and days leaves nine dates: $01\text{-}13,$ $01\text{-}31,$ $02\text{-}23,$ $03\text{-}11,$ $03\text{-}22,$ $10\text{-}13,$ $10\text{-}31,$ $11\text{-}03,$ and $11\text{-}30.$

Thus, the correct answer is E.