$\begin{array}{1 1}(A)\;100\\(B)\;104\\(C)\;108\\(D)\;110\end{array} $

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Let 2 be fixed at unit's place.The ten's place can be filled up in 6 ways.The hundred is place can be also be filled in 6 ways.

No of numbers that can be formed when 2 is at unit's place=$6\times 6=36$

Similarly when 4 is at unit's place,the number of numbers that can be formed =36

Again when 6 is at the unit's place.The number of numbers that can be formed =36

$\therefore$ The total numbers of ways when 3 digits numbers can be formed,the digits being repeated =$36\times 3=108$

Hence (C) is the correct answer.

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