Scenario I: The digits are \(2, 2, 2, 3.\) Total arrangements: \( \frac {4!}{3!} = 4 \)
Scenario II: The digits are \(2, 2, 3, 3.\) Total arrangements: \( \frac {4!}{2!.2!} = 6 \)
Scenario III: The digits are \(2, 3, 3, 3.\) Total arrangements: \( \frac {4!}{3!} = 4 \)
Scenario IV: The digits are \(2, 3, 3, 1.\) Total arrangements: \( \frac {4!}{2!} = 12 \)
Scenario V: The digits are \(2, 2, 3, 1.\) Total arrangements: \( \frac {4!}{2!} = 12 \)
Scenario VI: The digits are \(2, 3, 1, 1.\) Total arrangements: \( \frac {4!}{2!} = 12 \)
Therefore, the total number of possible arrangements is \( 12+12+12+4+6+4 = 50 \).