Question

The sum of all possible numbers that can be formed by using the digits \(2,3,5,7\) without repetition of digits is:

• A. \(17\times \dfrac{10^{4}-1}{9}\)
• B. \(33\times 34\times 101\)
• C. \(6\times \dfrac{10^{3}-1}{9}\)
• D. \(33\times 35\times 1001\)

Hint:

For numbers formed using all digits without repetition, each digit appears equally often in every place value. Use: \[ (n-1)! \] for the frequency of each digit in a particular place.

Answer 1

The Correct Option is B

Step 1: Find how many times each digit appears in each place.
Total 4-digit numbers = \(4! = 24\). Each digit (2,3,5,7) appears \(24/4 = 6\) times in each place (units, tens, hundreds, thousands).

Step 2: Compute the total sum.
Sum of digits \(= 2+3+5+7=17\). Contribution from each place \(= 6 \times 17 = 102\). Total sum \(= 102(1000+100+10+1) = 102 \times 1111 = 113322\). Now check: \(33 \times 34 \times 101 = 1122 \times 101 = 113322\).
\[\boxed{33 \times 34 \times 101}\]