$\begin{array}{1 1}(A)\;4000\\(B)\;4536\\(C)\;4530\\(D)\;3539\end{array} $

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- $nP_r=\large\frac{n!}{(n-r)!}$

Let the digits be 0 to 9

4-digit numbers are $10P_4$

This includes those number which have o in the beginning(one thousand's place)

3-digit numbers out of 9 digits 1-9 are $9P_3$

$\therefore$ 4 digit numbers which do not have zero in the beginning(on the extreme left)

$\Rightarrow 10P_4-9P_3$

$\Rightarrow \large\frac{10!}{(10-4)!}-\frac{9!}{(9-3)!}$

$\Rightarrow \large\frac{10!}{6!}-\frac{9!}{6!}$

$\Rightarrow \large\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{6\times 5\times 4\times 3\times 2\times 1}-\frac{ 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{6\times 5\times 4\times 3\times 2\times 1}$

$\Rightarrow 10\times 9\times 8\times 7-9\times 8\times 7$

$\Rightarrow 9\times 8\times 7[10-1]$

$\Rightarrow 9\times 9\times 8\times 7$

$\Rightarrow 4536$

Hence (B) is the correct answer.

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