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?
a) 420
b) 360
c) 320
d) 210
Answer: d
Explanation: 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
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