Приведём решение на языке Python.

from itertools import product
count = 0
for i in product("01x3x5x7x9xBxDEF", repeat = 4):
s = ''.join(i)
if s.count("E") == 1 and s[0] != "0":
if "Ex" not in s and "xE" not in s and "0E" not in s and "E0" not in s:
count += 1
print(count)

Ответ: 5336.

Приведём решение Сергея Донец на языке PascalABC.NET.

begin
(('0'..'9')+('A'..'F')).JoinToString.Cartesian(4)
.Where(s->s.CountOf('E')=1)
.Where(s->not s.IsMatch('[02468ACE]E'))
.Where(s->not s.IsMatch('E[02468ACE]'))
.Count(s->s[1] <> '0').Print;
end.