How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

3-digit numbers have to be formed using the digits 1 to 9.

Here, the order of the digits matters.

Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.

Therefore, required number of 3-digit numbers  $={ }^{9} \mathrm{P}_{3}=\frac{9 !}{(9-3) !}=\frac{9 !}{6 !}$

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