Question

The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is :

• A. 22264
• B. 26664
• C. 122234
• D. 122664

Hint:

Sum of numbers $= (\text{Sum of digits at unit place}) \times (111.......1)$.

Answer 1

The Correct Option is B

Step 1: Total permutations of $\{1, 2, 2, 3\} = \frac{4!}{2!} = 12$.
Step 2: Frequency of each digit at each place: Digit '1': Fix 1 at one place, others $\{2, 2, 3\}$ can be arranged in $\frac{3!}{2!} = 3$ ways. Digit '3': Fix 3 at one place, others $\{1, 2, 2\}$ can be arranged in $\frac{3!}{2!} = 3$ ways. Digit '2': Fix 2 at one place, others $\{1, 2, 3\}$ can be arranged in $3! = 6$ ways.
Step 3: Sum of digits at any place $= (1 \times 3) + (3 \times 3) + (2 \times 6) = 3 + 9 + 12 = 24$.
Step 4: Total sum $= 24(10^3 + 10^2 + 10^1 + 10^0) = 24 \times 1111 = 26664$.