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Permutations and Combinations
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 ?
$\begin{array}{1 1} 18 \\ 24 \\ 36 \\ 48 \end{array}$
jeemain
math
class11
ch7
permutations-and-combinations
permutations
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asked
Sep 3, 2013
by
rvidyagovindarajan_1
edited
Apr 5, 2016
by
meena.p
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1 Answer
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.)
answered
Sep 3, 2013
by
rvidyagovindarajan_1
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