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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?

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

Odd numbers can be arranged in $\large\frac{4!}{2!.2!}$ ways.
Even numbers can be placed in $\large\frac{3!}{2!}$ ways.
$\therefore$ The required no. of numbers = $\large\frac{4!}{2!.2!}.\frac{3!}{2!}$
$=6\times 3=18$