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

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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.

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