How many three-digit positive integers exist, all of whose digits are 2's and/or 5's and or 8's?
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)
+239 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.
