How many $4$-digit numbers are there with no digit repeated?

A four digit number is to be formed from the digits $0,1,2,3,4,5,6,7,8,9.$
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Since, leftmost place i.e thousand's place cannot have zero. So, there are $9$ ways to fill thousand's place.
Since, repetition is not allowed , so hundreds place can be filled by remaining $9$ digits .
So, hundred's place can be filled in $9$ ways.
Similarly, to fill tens place, we have $8$ digits remaining.
So,tens place can be filled by $8$ ways.
Similarly, to fill ones place, we have $7$ digits remaining.
So,ones place can be filled by $7$ ways.
So, required number of ways in which four digit numbers can be formed from the given digits is $9\times 9\times 8\times 7=4536$

Alternative Method:
The thousands place of the $4$-digit number is to be filled with any of the digits from $1$ to $9$ as the digit $0$ cannot be included. Therefore, the number of ways in which thousands place can be filled is $9$. 
The hundreds, tens, and units place can be filled by any of the digits from $0$ to $9$.
However, the digits cannot be repeated In the $4$-digit numbers and thousands place is already occupied with a digit. The hundreds, tens, and units place is to be filled by the remaining $9$ digits. 
Therefore, there will be as many such $3$-digit numbers as there are permutations of $9$ different digits taken $3$ at a time.
${}^9{P}_3=\frac{9!}{(9-3)!}=\frac{9!}{6!}$
Number of such $3$-digit numbers
 $\frac{9\times 8\times 7\times 6!}{6!}=9\times 8\times 7=504$
Thus, by multiplication principle, the required number of $4$-digit numbers is $9\times 504=4536$