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A door lock has 4 different numbers to choose from 1, 2, 3 and 4. To open the door one have to dial in a 6 digit code with the numbers mentioned. Repetitions are necessary and allowed.

How many possible combinations are there for this door lock?

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For every digit you have $4$ possibilities, so the total is:

$$4\cdot4\cdot4\cdot4\cdot4\cdot 4=4^6$$

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  • $\begingroup$ Yeah that's what i also thought, but my math teacher said it that the number was a lot higher than 4^6 = 4096 :/ $\endgroup$ – Agnar Holst Mar 20 '17 at 13:32
  • $\begingroup$ I'm afraid your teacher is wrong. If you can use just the digits $\{1,2,3,4\}$ and you must to place then in $6$ spots then the answer is $4^6$. Are you not missing any information? $\endgroup$ – Arnaldo Mar 20 '17 at 13:34
  • $\begingroup$ I will ask him in the next class and update the question if needed :) $\endgroup$ – Agnar Holst Mar 20 '17 at 13:37
  • $\begingroup$ @AgnarHolst: Imagine a smaller question. You can use just the digits $\{1,2\}$ and the code has $3$ digits. The possibilities are $\{(1,1,1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)\}$. It means $2^3$ possibilities. $\endgroup$ – Arnaldo Mar 20 '17 at 13:39

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