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?  Choose the correct option.
A) 420  B) 360  C) 320  D) 210  E) None
Solution:   Total number of arrangements = [8! /(4! *2! *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.   Answer: Option 210
<result>D</result>
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?