Question

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

Answer 1

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=$^9P_3$
$=\frac{9!}{(9-3)!}$

$=\frac{9!}{6!}$

$=\frac{9\times8\times7\times6}{6!}$

$=9\times8\times7=504$