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

Case I: The four digits are \(2, 2, 2, 3.\)
Total number of arrangements \(= \frac {4!}{3!} = 4\)

Case II: The four digits are \(2, 2, 3, 3.\)
Total number of arrangement \(= \frac {4!}{2!.2!} = 6\)

Case III: The four digits are \(2, 3, 3, 3.\)
Total number of arrangements \(=\frac {4!}{3!} = 4\)

Case IV: The four digits are \(2, 3, 3, 1.\)
Total number of arrangements \(= \frac {4!}{2!} = 12\) 

Case V: The four digits are \(2, 2, 3, 1.\)
Total number of arrangements \(= \frac {4!}{2!} = 12\)

Case VI: The four digits are \(2, 3, 1, 1.\)
Total number of arrangements \(= \frac {4!}{2!} = 12\)

So, total possible ways \(= 12+12+12+4+6+4 = 50\) ways.