I am a teacher. Subject: statistics How many combinations of three number codes can occur with a standard combination lock? The range of possible numbers is 0-39 which allows 40 different possible choices each turn of the dial. Each combination requires exactly three numbers (i.e. three turns of the dial). E.g. 35-40-5. Allow for reuse of the same number three times e.g.10-10-10. I don't think there is any preference shown for certain number combinations. Can you explain how the total is derived? Someone at the YMCA this week put this question to me and I gave an answer but I am not sure I am right--so I have come to you. logically it appears 40X40X40=64000. Actually on a standard combination lock it is clockwise to X , counterclockwise past X to Y (X cannot=Y and must be greater) clockwise to z (where Z is not equal to X nor Y). This changes how we view a combination lock beacuse X can be zero to 39 but cannot be 40 (since if it is 40 Y will not work), clockwise to Y where Y can be any number X or higher but not zero but can be 40, Z can be zero to 40 but cannot be either X nor Y. This seems to imply a maximum of 39X39X38=57798. What a half-ass guess on my part. John Hi John, I agree with your mathematical analysis in both cases, 40x40x40 and 39x39x38, but I am not sure exactly how the combination locks work. My recollection is this: The first number X can be any of the 40 numbers, 0-39. This gives 40 choices. Then you go counterclockwise, past X to Y which can be any number except X. This gives 39 choices. Now go clockwise again to the last number Z which can be any number except Y. This gives 39 choices. Thus I would get 40x39x39 possible combinations. I think you need to talk to someone who know how the locks work. Is 0-12-3 a valid combination? is 5-7-5 a valid combination? is 5-5-5 a valid combination? I don't know. Cheers, Penny Go to Math Central