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The number of numbers that can be formed by using digits 1,1,2,2,3,3,4, so that the odd digits always occupy odd places is?

$\begin{array}{1 1} 3!4! \\ 34 \\ 18 \\ 12 \end{array}$

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There are 4 odd places and 3 even places.
Odd numbers are 1,1,3,3.
Odd numbers can be arranged in $\large\frac{4!}{2!.2!}$ ways.
even numbers are 2,2,4, and even places are 3
Even numbers can be placed in $\large\frac{3!}{2!}$ ways.
$\therefore$ The required no. of numbers = $\large\frac{4!}{2!.2!}.\frac{3!}{2!}$
$=6\times 3=18$
answered Sep 3, 2013 by rvidyagovindarajan_1

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