Answer
58,663,725,120,000 strings
Work Step by Step
Total Number of arrangements of n objects if all are different = n!
If r! of them are same number of ways are given by $\frac{n!}{r!}$
because if the element that is repeated is exchanged with the similar element at some other position, it does not create a new outcome.
In our case we have two $0s$, four $1s$, three $2s$, one $3$, two $4s$, three $5s$, two $7s$, three $9s$ and the string length is 20.
So $\frac{20!}{2!.4!.3!.1!.2!.3!.2!.3!}$ = 58,663,725,120,000 strings