Question

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

Answer 1

Here total number of digits = 9

Number of digits used (no digit is repeated) = 3

$\therefore$ Number of permutations =${}^9{P}_3$

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