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.