Finding number of strings

Question

How many ternary strings of length 10 that contain exactly 2 1s and 3 2s. Since it must contain exactly 2 1s and 3 2s the other 5 spaces must be 0s. I was thinking it's 2!3!5! But apparently it's 10!/(2!3!5!) Can someone explain why we need the 10!

Answer 1

Out of the $10$ positions choose two of them to hold $1$, then out of the remaining $8$ positions choose three postions to hold $2$ and the remaining five positions are all $0$. Hence the number of ways is $$ \binom{10}{2}\binom{8}{3}\binom{5}{5}=\frac{10!}{2!3!5!}. $$