Combination lock probability

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Pingu

Junior Member
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Dec 22, 2005
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The dial on a three number combination lock contains markings to represent the numbers from 0 to 59. How many combinations are possible if the first and second numbers must be different, and the second and third numbers must be different?
 
There are 60 first numbers.
There are 59 second numbers.
There are 59 third numbers.
60*59*59.
 
What if the question was different, and it said:

The dial on a three number combination lock contains markings to represent the numbers from 0 to 59. How many combinations are possible if the first and second numbers must differ by at least three?
 
There are still 60 first numbers but only 55 second numbers. (if the first number were 0 then the second can be anything except 58, 59,0,1 or 2 eliminating 5 of the 60 possibilities)
PS If you don't consider 59 and 0 as differing by 1, that changes it. Let me know.
 
I don't think so. There are 60 numbers. If the first is 25 then that eliminates 23, 24, 25, 26 & 27 as the second. 60-5=55.
0 to 22 = 23 numbers
28 to 59 = 32 numbers
23+32=55. That checks.
25-22=3
28-25=3
They differ by at least 3.
What is your reasoning?
-----------------
Gene
 
I left open the possibility that 59 IS at least 3 from 0 though I said it wasn't. Depends on definitions. If that's the case you have to count 59 & 0, 58 & 1 as separate cases and see how many there are for each of them.
The other thing is you didn't specify the third number's spacing. That would affect the answer too. What answer are you trying for?
-------------------
Gene
 
Closest i've got is 198000
I went 60X55X60, it is wrong, the real answer is:
198360
 
OK, they do consider 59 to "differ by at least 3" from 0. Look at how many qualify for each of 0, 1, 58 and 59. Add those to 56*55 for the 2 to 57 range and it works.
(I wonder if they have ever watched the minute hand on a clock? I would say that 3:59 is within 3 minutes of 4:00 even though they say it isn't.)
-----------------
Gene
 
I'm sorry I understand
let's not start with the number 0
let's use number 10
 
But 10 is one of the 55 second numbers groups. The controversy is with 0, 1, 58 & 59.
Consider 0. Any number from 3 to 59 qualifies. 57 "good" numbers.
Do the same sort of count for 1, 58, and 59.
--------------
Gene
 
Shouldn't it be any number from three to fifty-eight?
I don't know how to do this question.
 
With first number 0: 0,1 & 2 are the banned numbers.
3 "bad", 57 "good."

With first number 1:
0, 1, 2 & 3 are the banned numbers.
4 "bad", 56 "good."

(Think about these and try for the high end with 58 & 59)

With first number 2:
0, 1, 2, 3 & 4 are the banned numbers and now we are into the 5 "bad", 55 "good" number groups ranging from 2 to 57. 56*55 "good."
-------------------
Gene
 
What don't you understand?
With first number 0: 0,1 & 2 are the banned numbers.
0-0 < 3. 0 is bad
1-0 < 3. 1 is bad
2-0 < 3. 2 is bad
3-0 = 3. 3 is good.
4-0 > 3. 4 is good.
.
.
57-0 > 3. 57 is good
58-0 > 3. 58 is good
59-0 > 3. 59 is good

3 "bad", 57 "good."
------------------
Gene
 
Bad = banned = numbers you can't use in the combination.
Good = numbers you can use in the combination.
---------------
Gene
 
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