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

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.

Number of such 3-digit numbers $={ }^{9} P_{3}=\frac{9 !}{(9-3) !}=\frac{9 !}{6 !}$

$=\frac{9 \times 8 \times 7 \times 6 !}{6 !}=9 \times 8 \times 7=504$

Thus, by multiplication principle, the required number of 4-digit numbers is

$9 \times 504=4536$