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!

The total numbers that can be formed using these $6$ digits is $6^3$. We find the number of numbers in which there is no $5$. Therefore numbers without the digit $ 5$ is $5^3$. Therefore the numbers with at least one $5$ in the number is $6^3-5^3=91$
Hence answer is $91$
Or you can do it the other way using cases: 
1) Number contains only one $5$ and all letters are different:
Number of such numbers = $^5C_2. 3!=60$
2) Numbers with only one $5$ and other letters alike:
Number of such numbers=$^5C_1.\frac {3!}{2!}=15$
3)Numbers with two $5's$ and one other number:
Number of such numbers=$^5C_1.\frac {3!}{2!}=15$
4)Numbers with three $5's= 1$
Hence the total number of numbers satisfying the given condition =$60+15+15+1=91$.
