License Plate Combinations: Clarifying the Math

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The discussion revolves around calculating possible license plate combinations using letters and digits. One participant initially suggested a formula based on fixed positions for letters and digits, while another introduced permutations to account for different arrangements. The conversation highlighted the importance of understanding the problem's context, particularly regarding which letters are valid on license plates, as certain letters may be excluded in real-world scenarios. Participants debated the correct mathematical approach and clarified the formula for permutations, ultimately converging on a consensus regarding the total number of valid combinations. The thread concluded with a recognition of the complexities involved in license plate regulations and the assumptions made in the calculations.
  • #31
DaveC426913 said:
The set used for license plates, of course. i.e.: 26, sans I, O and Q.
That was an additional assumption/restriction on your part, not one that was given in the problem statement, and one that likely would have led to an incorrect answer.
 
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  • #32
DaveC426913 said:
While that's probably outside the scope of the question
Yes.
DaveC426913 said:
mentioning it, and providing a second solution - might net you bonus marks!
Or not...
DaveC426913 said:
But the problem does specify a context (license plates) - one in which a non-conventional letter set is specified.
No, there is no mention whatsoever in the problem statement (which I quoted several posts back) about any non-conventional letter sets or any restrictions other than three letters followed by three digits.

Since the original question has been answered to the OP's satisfaction, and further talk about reduced sets of letters is off-topic, I am closing this thread.
 
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  • #33
requied said:
Of course I can, here below;
View attachment 263365
I hope our answers are the same 😅
You are double counting some of the numerics. E.g. you are counting 112 as an example of x12 and as an example of 1x2.
 
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  • #34
haruspex said:
You are double counting some of the numerics. E.g. you are counting 112 as an example of x12 and as an example of 1x2.
So the answer became 36.504
 
  • #35
requied said:
So the answer became 36.504
Yes.
 
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  • #36
haruspex said:
Yes.
I guess there is no a similar application at letters order.
 
  • #37
requied said:
I guess there is no a similar application at letters order.
Quite so, that was easier. Had you only been told there was at least one A then, depending on your approach, there would have been a risk of double counting.
(But in that specific case the easy way is 263-253.)
 
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