Solve Permutation Problem: 36 Combination Lock

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A 3 combination lock with 36 possible numbers allows for combinations where numbers can be reused but not consecutively. The correct calculation for the total number of combinations is 36 (for the first number) multiplied by 35 (for the second number, since it cannot be the same as the first) and again 35 (for the third number, which cannot be the same as the second). This results in a total of 44,100 different possible combinations. The focus of the discussion is on understanding the correct mathematical approach rather than just obtaining the answer. The participants confirm the calculation method as accurate.
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1. I've only just scrapped through math in high school but recently I have started to take up an interest in the subject. I was listening to a math podcast and they posted a question to be answered the following episode. However, I cannot seem to track that one down and I really need an answer because I'm starting to obsess on it.



2. A 3 combination lock has 36 possible numbers. What are all the possible combinations if you can reuse numbers but not consecutively? Examble: 121 but not 112, or 122



3. I assumed that you would take 36*35*35 which would be 44,100 different possible combinations. Is this correct? I'm more interested in the correct equation to use, then I am the answer.
 
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welcome to pf!

hi ajgrebel! welcome to pf! :wink:
ajgrebel said:
2. A 3 combination lock has 36 possible numbers. What are all the possible combinations if you can reuse numbers but not consecutively? Examble: 121 but not 112, or 122

3. I assumed that you would take 36*35*35 which would be 44,100 different possible combinations. Is this correct? I'm more interested in the correct equation to use, then I am the answer.

yes that's correct :smile:
 
Thank you for letting me know
 

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