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How many 9 digit numbers can be formed from the numbers 223355888 so that the odd digits occupy even positions.

$\begin{array}{1 1} 16 \\ 36 \\ 60 \\ 180 \end{array}$

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There are 4 even places and 5 odd places.
Odd digits are 3355.
Odd numbers are placed in these 4 places in $\large\frac{4!}{2!.2!}$ ways.
Even digits are 22888.
These even digits are placed in 5 odd places in $\large\frac{5!}{3!.2!}$ ways.
$\therefore $ No. of required 9 digit numbers= $\large\frac{4!}{2!.2!}.\large\frac{5!}{3!.2!}$
answered Aug 14, 2013 by rvidyagovindarajan_1

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