Question

A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

Answer 1

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 \( 12+12+12+4+6+4 = 50 \).