How many arrangement of four 0 zeroes two 1 and two 2 are there in which the first 1 occur before the first 2?
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Step-by-step explanation:
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.
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