N n of hexadecimal digits, and the number of all numerical number, the combination of 50% should be how to calculate?

Example 1:

The 2 decimal digits, and the number of all numerical combination of 50%, there are about 100, only one kind of the combination way: end combination:
10,11,12,13 00,01,,,,,,,, 99



Example 2:

The three decimal digits, and the number of all numerical combination of 50% the number of about 87, the combination of n, or the way the kinds:
"Nine minus,"
"Six,0,2,"
"Six,9,6,"
"Five,8,8,"
"Six,8,9,"
"Five,6,1,"
"Seven,4,9,"
"3,9,2",
"2,0,6,"
7 and 4 ", "
"0, 6,"
"Five,1,9,"
"Eight,7,7,"
"Four,3,5,"
"0,9,8,"
"Seven,8,5,"
"1,3,9",
"2,4,8,"
"Seven,2,8,"
"Eight,9,3,"
"2,6,9,"
"Seven,7,1,"
"0,8,6,"
"1,4,2",
"2,9,1,"
"Eight,6,6,"
"Five,0,3,"
"3, 2, 3,"
"Five,9,4,"
"1,6,0",
"5 0,"
"9,5,8,"
"4 and 6,"
"Six,7,0,"
"9,7,9,"
"9,3,7,"
"3,8,7",
"Eight, four, five,"
"0,5,5,"
"A proposal,"
"Six,1,5,"
"Eight,3,8,"
"4, 2, 1,"
"0,7,3,"
"Seven,6,2,"
"2,5,3,"
"Seven,9,7,"
"Eight,1,2,"
"1,0,5",
"Five,5,2,"
31 "2",
"3,7,5",
"Four,7,2,"
"1,2,6",
"3,0,4",
"3,4,1",
"0, 0,"
"2,7,4,"
"Eight 4-trichlorobenzene,"
"9,8,3,"
"1,1,1",
"0,3,1,"
6, 3 ", "
"Four,6,8,"
"9, 0,"
"2,2,5,"
"Eight,8,1,"
"2,1,7,"
"9,9,5,"
"3,3,6",
"Six,6,4,"
"Four,1,3,"
"Seven,5,0,"
"1,8,4",
"Four,4,7,"
"9, 1,"
"3, 9,"
"0,4,4,"
"Four,0,9,"
"Four,8,0,"
"1, 5, 7,"
"3,1,8",
"0,2,2,"
"Seven,1,6,"
"Six,2,7,"
"Five,7,6,"
"Six,5,1
"

Example 3:

The 4 decimal digits, and the number of all numerical combination of 50% the number of about 750, combination of n, or the way the kinds:



The sample n:
20 of 20 hexadecimal digits, and the number of all numerical different reach about 50% of the group? A:

For the formula?