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The number of 4 digit numbers divisible by 3 that can be formed by 4 different even digits (0,2,4,5,6,8) is ?

$\begin{array}{1 1} 18 \\ 24 \\ 36 \\ 48 \end{array}$

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Since the number is divisible by 3, the sum of the digits should be multiple of 3.
$\therefore$ The digits can be $0,2,4,6\: or \: 0,4,6,8$
Hence the required no. of numbers = $(4!-3!)+(4!-3!)=36$
(0 cannot come in $1^{st}$ position.)
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