Question 1-2.

Using only the digits 1, 2 and 3, to form a six-digit numbers,
there are total 3^6 of possibilities = 729
In all of them, there are 2^6 numbers do not contain of 3 (i.e. only 1 and 2)
that is 64 possibilities.
Therefore, the six-digit numbers form from 1,2 & 3 only,
there are 729 - 64 = 665 numbers contain at least one 3.

Using only the digits 1, 2 and 3, how many different six-digit numbers can be formed that contain at least one 3 ?

3*3*3*3*3*1/3=81 the first to fifth 3 =any number(1-3), the 1/3 is the chance for 3
81different six-digit numbers can be formed that contain at least one 3