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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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In a certain city,all telephone numbers have six digits,the first two digits always being 41 or 42 or 46 or 62 or 64.How many telephone numbers have all six digits distinct.

$\begin{array}{1 1}(A)\;8200\\(B)\;8300\\(C)\;8400\\(D)\;8500\end{array} $

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1 Answer

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  • $P(n,r)=\large\frac{n!}{(n-r)!}$
Total no of digits in the telephone number =6
Given :
Possible first two digit =41,42,46,62 or 64
$\Rightarrow 5$
First two digits can be filled in 5 ways
The remaining four digit can be filled in =$8P_4$ ways
$8P_4=\large\frac{8!}{(8-4)!}$
$\Rightarrow \large\frac{8!}{4!}=\frac{8\times 7\times 6\times 5\times 4!}{4!}$
$\Rightarrow 1680$
Total no of ways =$5\times 1680$
$\Rightarrow 8400$
Hence (C) is the correct answer.
answered May 15, 2014 by sreemathi.v
 

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