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The sum of the digits in unit place of all the numbers formed using 3,4,5,6 taking all at a time is?

$\begin{array}{1 1} 18 \\ 36 \\ 108 \\ 432 \end{array}$

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Fixing 3 in unit place there are $3!=6$ numbers.
Fixing 4 in unit place there are $3!=6$ numbers
Similarly by fixing 4 and 5 in unit place we get 6 numbers in each
$\therefore$ Each of the digits 3,4,5,6 come in unit place in 6 numbers
$\therefore$ The sum of the digits in unit place of the the numbers=
$6\times (3+4+5+6)=108$
answered Aug 15, 2013 by rvidyagovindarajan_1
 

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