What is the correct answer ?

0,1,2,3 occur atleast once..

now they in themselves have a sum of 6

sum of the digits is 10

so the remaining 2 digits need to make up for 4 more

so the ways are

0 and 4
1 and 3
2 and 2

Find the sum with these three cases :)

done :)

first calculation gives 530.. but i will recheck..

case 1 .. . 0,0,1,2,3,4

So the total no of ways is 4.5!/2!

case 2... 0,1,1,2,3,3

first digit 1: 5!/2!
first digit 2: 5!/(2!.2!)
first digit 3: 5!/2!

case 3... 0,1,2,2,2,3

first digit 1: 5!/3!
first digit 2: 5!/2!
first digit 3: 5!/3!

so the answer is 490.. i made a slight calc mistake in taking 3! as 3 :P