Discussion

Find the sum of all four digit numbers that can be formed by the digits 1, 3, 5, 7, 9 without repetition.

Answer: B

Four-digit numbers from five distinct digits, no repetition:

\[ {}^{5}P_{4} = 120 \text{ numbers} \]

By symmetry each digit occupies each position equally often:

\[ \frac{120}{5} = 24 \text{ times per position} \]

The digits sum to \( 1+3+5+7+9 = 25 \), so each position contributes \( 24 \times 25 = 600 \) to its place value.

\[ \text{sum} = 600 \times (1000 + 100 + 10 + 1) = 600 \times 1111 \]

\[ = 666{,}600 \]

The symmetry argument avoids listing 120 numbers: no digit is favoured in any column.

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