So the question is: How many $6$ digit numbers can be formed with the numbers $0, 1, 2$ if the number must contain at least one $0$?
I have searched and found some similar questions so I hope this is not a duplicate. I would please like you to explain where my thinking goes wrong with this problem.
My #1 attempt at a solution:
To start, to create a number that satisfies the question it has to be on the form:
$ 1 \times 3 \times 3 \times 3 \times 3 \times 3 $ Where the "$1 \times$" represent that there must be at least one $0$.
This gives us $3^5$ possibilities.
Since the "mandatory" $0$ could be placed in six different spots we get that there are: $6 \times 3^5$ possibilities.
This leaves us with a set of numbers that have duplicates and I cannot see any direct formula to subtract these duplicates.
My #2 attempt at a solution:
Here I can start with a set of all the possible numbers that could be generated:
$ 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$ and subtract that with the set of numbers that does not contain a zero: $ 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$.
This would give me the answer: $3^6 - 2^6$. Which seems to be wrong.
Please help me out here.