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 determine the count of four-digit numbers composed solely of digits 1, 2, and 3, where both 2 and 3 are present at least once, we follow these steps:

• Step 1: Determine the total number of possible four-digit numbers:
  • Each of the four positions can be filled with one of three digits (1, 2, or 3).
  • Total combinations: $3^4 = 81$
• Step 2: Calculate the number of cases where either 2 or 3 is absent:
  • Numbers formed using only 1 and 3 (no 2): $2^4 = 16$
  • Numbers formed using only 1 and 2 (no 3): $2^4 = 16$
  • Numbers formed using only 1 (no 2 and no 3): $1^4 = 1$
  • Applying the Principle of Inclusion-Exclusion to find cases where at least one of 2 or 3 is missing: $16 + 16 - 1 = 31$
• Step 3: Subtract excluded cases from the total:
  • Numbers that contain both 2 and 3 at least once: $81 - 31 = 50$
• Step 4: State the final result:
  • The count of four-digit numbers using only digits 1, 2, and 3, with both 2 and 3 appearing at least once, is 50.

Conclusion: The desired quantity is 50.