Question

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

Answer 1

Finding $3-$ digit numbers when no-digit is repeated
Total number of digits $\left(n\right)=9$
Given, $3$ digit number $\Rightarrow r=3$
$\therefore$ Required number of ways
$={}^n{P}_r=\frac{n!}{(n-r)!}=\frac{9!}{(9-3)!}=\frac{9!}{6!}=\frac{9\times 8\times 7\times 6!}{6!}$
$=9\times 8\times 7=504$.