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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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The number of natural numbers less than 1000 in which no two digit are repeated is

$\begin{array}{1 1}(A)\;738\\(B)\;792\\(C)\;837\\(D)\;720\end{array} $

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1 Answer

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  • $P(n,r)=\large\frac{n!}{(n-r)!}$
Total number of 1 digit number =9
The number of 2 digit non-repeated number =$9\times 9$
$\Rightarrow 81$
The number of 3 digit non-repeated number =$9\times 9P_2$
$\Rightarrow 9\times 9\times 8$
$\Rightarrow 648$
$\therefore$ Required number of ways =9+81+648
$\Rightarrow 738$
Hence (A) is the correct answer.
answered Jun 20, 2014 by sreemathi.v
wow wow wowo wow ow owo wow ow owow wo wow wo wo wow wo w
 

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