Huge number, right?

$4=1+1+1+1 \\ =2+2 \\ =1+1+2 \\ =1+3 \\ = 4$

For each case put first digit as non-zero and then count the number of ways.

Case(1) : Putting 1 as first digit there are remaining 9 digits( 3 1's and 6 zeroes)

So total ways : $\dfrac{9!}{6! 3!}=84$

Case(2) : Put 2 as first digit.Remaning digits are 1 2's and 8 zeroes.

So total ways : $\dfrac{9!}{8!}=9$

Case(3) : It has two subcases first put 1 as first digit and then put 2 as second digit.in first case putting 1 as first digit remaining digits are 1 1's,1 2's and 7 zeroes.So total ways are $\dfrac{9!}{71}=72$

Now put 2 as first digit.The remaining digits are 2 1's and 7 zeroes.So, total no of ways are $\dfrac{9!}{7! 2!}=36$

Case(4) : In this case put 1 and putting 3 are similar case.So I will calculate only one way.putting 1 as first digit remaining digits are 1 3's and 8 zeroes.So total number of ways are $\dfrac{9!}{8!}=9$

So total ways $9 \times 2=18$

Case(5) : In this case Put 4 as first digit and remaining digits are 9 zeroes.This has only $1$ way.

So total number of ways are $84+9+72+36+18+1= \boxed{220}$