A) 26664
B) 39996
C) 38664
D) none of these
Correct Answer: C
Solution :
[c] The number of numbers with 0 in the unit's place is 3!=6. The number of numbers with 1 or 2 or 3 in the unit's place is 3!-2!=4. Therefore, the sum of the digits in the unit's palace is \[6\times 0\ +4\times 1+4\times 2+4\times 3=24.\] Similarly, for the ten?s and hundred's places, the number of numbers with 1 or 2 in the thousand's place is 3! Therefore, the sum of the digits in the thousand's place is \[6\times 1+6\times 2+6\times 3=36.\] Hence, the required sum is \[36\times 1000+24\times 100+24\times 10+24.\]
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