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?

Answer: Option D

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