Solve Permutation Problem: 36 Combination Lock

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SUMMARY

The discussion centers on calculating the number of possible combinations for a 3-digit combination lock with 36 numbers, allowing for number reuse but prohibiting consecutive repetitions. The correct formula for this scenario is 36 * 35 * 35, resulting in 44,100 unique combinations. Participants confirmed this calculation and emphasized the importance of understanding the underlying equation rather than just the final answer.

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  • Understanding of basic combinatorial mathematics
  • Familiarity with the concept of permutations and combinations
  • Knowledge of restrictions in combinatorial problems
  • Basic arithmetic skills for multiplication
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  • Study combinatorial mathematics to deepen understanding of permutations and combinations
  • Explore problems involving restrictions in combinations, such as consecutive number limitations
  • Learn about advanced combinatorial techniques and their applications
  • Practice solving similar problems using different numbers of digits and constraints
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This discussion is beneficial for mathematics enthusiasts, students learning combinatorial concepts, and anyone interested in solving practical problems related to combinations and permutations.

ajgrebel
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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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