How many arrangements of four 3’s two 1’s and two 2’s are there in which 1 occurs before 2?

210

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420

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105

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315

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Solution

The correct option is A 210
Total Number of arrangements
$= \frac{∠ 8}{∠ 4 ∠ 2 ∠ 2} = 420$
Since there are two 1 ‘s, two 2’s, the number of arrangements say x in which 1 occurs before 2 is same as the 2 occurs before $1. 2 x = 420 \Rightarrow x = 210$

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