Question

The sum of all the 4-digit numbers formed by taking all the digits from \(2, 3, 5, 7\) without repetition is

• A. \(331122\)
• B. \(123312\)
• C. \(113322\)
• D. \(132132\)

Hint:

To compute the sum of all permutations, use symmetry and consider the contribution of each digit separately.

Answer 1

The Correct Option is C

Step 1: Calculating the number of permutations.
Since we are forming 4-digit numbers using all four digits without repetition, the total number of such numbers is: \[ 4! = 24 \] Step 2: Contribution of each digit to sum.
Each digit appears in each place (thousands, hundreds, tens, units) exactly: \[ \frac{4!}{4} = 6 \text{ times} \] Thus, the total sum contributed by a single digit in all positions is: \[ 6 \times (1000 + 100 + 10 + 1) = 6 \times 1111 = 6666 \] Step 3: Computing total sum.
Summing over all digits: \[ (2+3+5+7) \times 6666 = 17 \times 6666 = 113322 \] Thus, the correct sum is: \[ \mathbf{113322} \]