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How many three digit numbers that are divisible by 3 can be formed using the digits 0,1,2,3,4,5 without repeating the digits?

$\begin{array}{1 1} 100 \\ 60 \\ 40 \\ 24 \end{array}$

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For a 3 digit number to be divisible by 3, sum of its digits
should be divisible by 3.
With 0 as one of the digits the remaining 2 digits can be (1,2),(1,5),(2,4),(4,5)
After choosing the three numbers they can be arranged in 3! ways.
But with 0 as one of the digits, 0 cannot come in hundreds place for 3 digit numbers..
$\therefore$ No. of numbers with 0 as one of its digits = $4\times (3!-2!)=16$
With 1 as one of the digits the remaining 2 digits can be $ (2,3),(3,5)$
$\therefore$ No. of numbers with 1 as one of its digits $=2\times 3!=12$
With 2 as one of the digits the remaining 2 digits are (3,4)
$\therefore $No. of numbers with 2 as one of its digits =$1\times 3!=6$
With 3 as one of the digits the remaining 2 digits are (4,5)
$\therefore$No. of numbers with 3 as one of its digits $= 1\times 3!=6$
Total numbers $= 16+12+6+6=40$
answered Aug 14, 2013 by rvidyagovindarajan_1
 

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