How many different $7$-digit numbers can be formed from $0,1,2,2,3,3,3$ assuming no number can start with $0$? How many numbers will end with $0$?

Question

I got this question and wanted to confirm my solution.

How many different $7$-digit numbers can be formed from $0,1,2,2,3,3,3$ assuming no number can start with $0$? How many numbers will end with $0$?

For the first part, I got $360$, and for part 2, I got $60$.

Did I do this correctly?

Answer 1

Yes, those answers are correct. The answer for $1$) is

$$ \frac{7!-6!}{3!\cdot2!}=360\;, $$

and the answer for $2$) is

$$ \frac{6!}{3!\cdot2!}=60\;. $$