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

Correct Answer: 50

To solve this problem, we need to count all possible four-digit numbers formed using only the digits 1, 2, and 3, with both 2 and 3 appearing at least once. Let's break down the solution: 

• Step 1: Calculate total numbers formed:
  • Each digit has 3 possibilities: 1, 2, or 3.
  • Total possible numbers = $3^4 = 81$
• Step 2: Calculate cases where 2 or 3 do not appear:
  • Case without 2: Use only 1 and 3 → $2^4 = 16$
  • Case without 3: Use only 1 and 2 → $2^4 = 16$
  • Case without both 2 and 3: Use only 1 → $1^4 = 1$
  • By Inclusion-Exclusion Principle: 
    Total = $16 + 16 - 1 = 31$
• Step 3: Subtract to get cases with both 2 and 3:
  • Valid cases = $81 - 31 = 50$
• Step 4: Final Answer:
  • The number of four-digit numbers using only 1, 2, 3 with both 2 and 3 appearing at least once is 50.

Conclusion: The required number is 50.