It does take some time to digest

0+1+2+3=6 so the remaining 2 digits must add up to 4 making them either 0 and 4, 1 and 3 or 2 and 2. With 0 and 4 we are looking for the number of different orders for the digits 0, 0, 1, 2, 3 and 4. 0 cannot go at the beginning to so this is4*5!/2!=240

Next we need the possible combinations of 0, 1, 1, 2, 3 and 3. Again we cannot start with 0 so the number is 5!/2!+5!/2!+5!/2!*2!=150

Next is the combinations for 0, 1, 2, 2, 2, 3 which will be 5!/3!+5!/3!+5!/2!=100

Adding these numbers gives 490 which should be the right answer.

It is very easy to see that: the number can be built from three sets of digits: i) 0 0 1 2 3 4; ii) 0 1 1 2 3 3; iii) 0 1 2 2 2 3 So the number of 6 digits number is: $\frac{6!}{2}-5!+\frac{6!}{2^2}-\frac{5!}{2^2}+\frac{6!}{3!}-\frac{5!}{3!}=\boxed{490}$