How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?

Total number of arrangements = $$\frac{{8!}}{{4! \times 2! \times 2!}}$$   = 420

Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements = 210