CAT Exam 2021 Slot 3 Solved Paper

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 : $\frac{4!}{3!}=4$ Case 2: The four digits are $2, 2, 3, 3$. The total number of four-digit numbers formed using this are : $\frac{4!}{2! \cdot 2!}=6$ Case 3: The four digits are $2, 3, 3, 3$ . The number of possible 4 digit numbers are : $\frac{4!}{3!}=4$ Case 4: The four digits are $2, 3, 3, 1$. The number of possible 4 digit numbers are : $\frac{4!}{2!}=12$ Case 5: Using the digits $2, 2, 3, 1$. The number of possible 4 digit numbers are : $\frac{4!}{2!}=12$ Case 6: Using the digits $2, 3, 1, 1$. The number of possible 4 digit numbers are : $\frac{4!}{2!}=12$ A total of $12 + 12 + 12 + 4 + 6 + 4 = 50 \text{ possibilities}$.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- $\frac{4!}{2!} \times 3= 36$ ( When one digit is used twice and the remaining two once ) $\frac{4!}{3!} \times 2= 8$ ( When 1 is used 0 times and 2 and 3 is used 3 times or 1 time ) $\frac{4!}{2! \cdot 2!} = 6$( When 2 and 3 is used 2 times each ) So total numbers $= 36 + 8 + 6 = 50$