Explanation

(c)
The sum of all the four digit numbers using the digits 3, 5, 7 and 9
$= (3 + 5 + 7 + 9) \times (4 - 1)!\left( \frac{10^{4} - 1}{10 - 1} \right)$
$= 24 \times 6 \times \left( \frac{10^{4} - 1}{10 - 1} \right)$
$= \frac{24 \times 6 \times 9999}{9}$
$\therefore$ Required average =$\frac{24 \times 6 \times 9999}{9 \times 24} = 6666$