Question

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

Answer 1

Here total number of digits = 10

Number of digits used (no digit is repeated) = 4

Since 0 cannot be filled in the fourth place, number of permutations for fourth place = 9

Now the remaining three places can be filled with 9 digits.

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

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

Hence total number of permutations = $9\times 504=4536.$