There are [tex]\dbinom{10}2[/tex] ways of picking 2 of the 10 available positions for a 0. 8 positions remain.
There are [tex]\dbinom83[/tex] ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's [tex]\dbinom55=1[/tex] way of doing that.
So we have
[tex]\dbinom{10}2\dbinom83\dbinom55=\dfrac{10!}{2!(10-2)!}\dfrac{8!}{3!(8-3)!}\dfrac{5!}{5!(5-5)!}=2520[/tex]
The last expression has a more compact form in terms of the so-called multinomial coefficient,
[tex]\dbinom{10}{2,3,5}=\dfrac{10!}{2!3!5!}=2520[/tex]
