The question asks for the number of 4 digit numbers using only the digits 1, 2, and 3 such that the digits 2 and 3 appear at least once.
The different possibilities include :
Case 1: The four digits are $( 2, 2, 2, 3)$. Since the number 2 is repeated 3 times. The total number of arrangements are :
${4!}/{3!}=4$
Case 2: The four digits are $2, 2, 3, 3$. The total number of four-digit numbers formed using this are : ${4!}/{2!.2!}=6$
Case 3: The four digits are $ 2, 3, 3, 3$ . The number of possible 4 digit numbers are : ${4!}/{3!}=4$
Case 4: The four digits are $2, 3, 3, 1$. The number of possible 4 digit numbers are : ${4!}/{2!}=12$
Case 5: Using the digits $2, 2, 3, 1$. The number of possible 4 digit numbers are : ${4!}/{2!}=12$
Case 6: Using the digits $2, 3, 1, 1$. The number of possible 4 digit numbers are : ${4!}/{2!}=12$
A total of $12 + 12 + 12 + 4 + 6 + 4 = 50 \possibilities$.Alternatively We have to form 4 digit numbers using $1,2,3$ such that $2,3 $ appears at least once
So the possible cases :
Now we get- ${4!}/{2!} × 3= 36$ ( When one digit is used twice and the remaining two once )
${4!}/{3!} × 2= 8$ ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time )
${4!}/{2!.2!} = 6$( When 2 and 3 is used 2 times each )
So total numbers $= 36 + 8 + 6 =50$