Question

How many 3-digit numbers using permutation can be formed from the digits 1, 2, 3, 4 and 5 assuming that repetition of the digits is allowed?

Answer 1

3-digit number using permutation 135

Answer 2

Answer = 125

We are assuming that the digits can be repeated, so:

On 1st position, 3 digit numbers can exist in = 5 ways

Likewise, On 2nd position, 3 digit numbers can exist in = 5 ways

And in the same way, On 3rd position, 3 digit numbers can exist in = 5 ways

So, 1st way x 2nd way x 3rd way = 5x5x5 = 125