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!
 A: There is a single number with three $5$, there are $3\times 5^2$ numbers with a single $5$, and $3\times 5$ numbers with two $5$. It makes $91$ numbers.
A: You are overcounting. 
Do not include the permutations of $5$s among themselves. Hence, divide into cases and solve.
The number of ways in which 5 can be filled in the three places is $3$.
For one $5$: 
$3\times 5\times 5= 75$ //No repetition problems here. 
For two $5$s :
$\dbinom32 \times 5 = 15$ 
For all $5$s :
Clearly there's only 1 possibility: $555$. 
Thus, the required answer is $75+15+1= 91$
Edit:
Numbers over-counted by you: 
$ 1)255$
$2)552$
$3)525$
$4)355$
$5)553$
$6)535$
$7)455$
$8)554$
$9)545$
$10)655$
$11)556$
$12)565$
$13)855$
$14)558$
$15)585$
$16)555 $
$17)555$
Hence,you get a difference of $108 -91= 17$ in your answer. 
A: 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 =$60+15+15+1=91$.
A: Take all the possible numbers with $5$
then substract all the numbers without $ 5$
Result: $6^3-5^3=91$
