I still doubt that the OP means to ask the question that implies that order makes a difference.

This is a well known question about integer partitions.

Look at this webpage.
Now my subscription is for WolframalphaPro. So I hope all can see that there are 42 partitions of 10.

But for this problem, using only single digits there would be only 41.

Here they are:

If from that list we choose $3+2+2+1+1+1=10$ and order makes a difference, then there are $\dfrac{6!}{(2!)(3!)}=60$ in that count.

So we would need to do that for each of the 41 cases.