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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that odd digits occupy even ositions

$\begin{array}{1 1}(A)\;16\\(B)\;36\\(C)\;60\\(D)\;180\end{array} $

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  • $n!=n(n-1)(n-2)(n-3).......(3)(2)(1)$
There are 5 odd places
These can be filled by using 2,2,8,8,8 in $\large\frac{5!}{2!3!}$
$\Rightarrow \large\frac{5\times 4\times 3!}{2\times 3!}$$=10$ ways
A even places can be filled by using 3355 in =$\large\frac{4!}{2!\times 2!}$
$\Rightarrow \large\frac{4\times 3\times 2!}{2\times 2!}$
$\Rightarrow 6$
$\therefore$ Total no. of numbers =$10\times 6=60$
Hence (C) is the correct answer.
answered May 20, 2014 by sreemathi.v
 

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