Let $$N$$ be the number of distinct 8-digit numbers obtained by arranging the six numbers $$0, 1, 2, 3, 10, 23$$, where the first digit of the 8 digit number is not zero. Find the sum of the digits of $$N$$.

Treating the six given numbers as blocks, there are $$6! - 5! = 600$$ arrangements not beginning with 0, but different arrangements can produce the same 8-digit number. Counting each number once through the representation in which the block 10 is the first occurrence of 10 and the block 23 is the first occurrence of 23, the arrangements breaking these rules number $$\frac{5!}{2} = 60$$ and $$\frac{5! - 4!}{2} = 48$$, with $$\frac{4!}{4} = 6$$ restored by inclusion and exclusion. Hence $$N = 600 - 60 - 48 + 6 = 498$$ and the digit sum is $$4 + 9 + 8 = 21$$.