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.

Thank you!

  • 1
    $\begingroup$ Your second attempt is very close, but you will find that the product of threes has to start with a $2$, since the first digit cannot be a $0$ (it wouldn't be a $6$-digit number if it did). So $2\cdot 3^5 - 2^6$ would be my guess. $\endgroup$ – Arthur Jan 21 '14 at 10:18

Your second solution is almost right, but $3^6$ includes the numbers which have $0$ as the left-most digit. So, you need to eliminate these numbers. Hence, the answer is $$2\cdot 3^5-2^6.$$

  • $\begingroup$ Thank you! It all seems so clear now :)! Easy to get lost in the "abstraction"! $\endgroup$ – Lukas Arvidsson Jan 21 '14 at 12:16
  • $\begingroup$ You are welcome! $\endgroup$ – mathlove Jan 21 '14 at 12:17

The number of all possible numbers is not $3^6$ as the first digit must be $1$ or $2$, so the number of all possible $6$ digit numbers is $2\cdot 3\cdot 3\cdot 3\cdot 3\cdot 3$

  • $\begingroup$ The number must contain a zero ... $\endgroup$ – Michael Hoppe Jan 21 '14 at 10:43
  • $\begingroup$ @MichaelHoppe I edited the answer to be more clear in what I wanted to say. It was ambiguous at first. $\endgroup$ – 5xum Jan 21 '14 at 10:46
  • $\begingroup$ Thank you for your answer! $\endgroup$ – Lukas Arvidsson Jan 21 '14 at 12:17

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