How many $3$-digit numbers can be formed using the digits $ 2,3,4,5,6,8 $ such that the number contains the digits $5$ and repetitions are allowed? The solution I have is by counting the complement which gives an answer of $91$. But I think that it should be solved as follows-
Let the $3$-digit number be denoted by $3$ boxes. We can put the digit $5$ in one of the three boxes and we can put fill remaining two boxes with $6$-digits. Hence answer is $6\cdot 6\cdot 3= 108$. Help!

Take all the possible numbers with $5$
then substract all the numbers without $ 5$
Result: $6^3-5^3=91$
