# How many n-digit numbers are there with no digit repeated?

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

Toolbox:
• $nP_r=\large\frac{n!}{(n-r)!}$
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.