A) 4444
B) 5555
C) 6666
D) 7777
E) 8888
Correct Answer: C
Solution :
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 \[=\frac{24\times 6\times 9999}{9\times 24}=6666\]
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