How many three-digit positive integers exist, all of whose digits are 2's and/or 5's and or 8's?

Guest Jan 3, 2021

#1**+1 **

Total Permutations =3^3 = 27

{2, 2, 2} | {2, 2, 5} | {2, 2, 8} | {2, 5, 2} | {2, 5, 5} | {2, 5, 8} | {2, 8, 2} | {2, 8, 5} | {2, 8, 8} | {5, 2, 2} | {5, 2, 5} | {5, 2, 8} | {5, 5, 2} | {5, 5, 5} | {5, 5, 8} | {5, 8, 2} | {5, 8, 5} | {5, 8, 8} | {8, 2, 2} | {8, 2, 5} | {8, 2, 8} | {8, 5, 2} | {8, 5, 5} | {8, 5, 8} | {8, 8, 2} | {8, 8, 5} | {8, 8, 8} (total: 27)

Guest Jan 3, 2021

#2**0 **

If you decide to list all of them,

You would get 27(As Guest said)

But there is an easier way,

The first digit has three possibilities; it can either be a 2 a 5 or an 8.

The second digit has three possibilities; it can either be a 2 a 5 or an 8.

The third digit has three possibilities; it can either be a 2 an 8 or a 5.

So \(3*3*3=27 \) possible integers.

DewdropDancer Jan 3, 2021