License Plate Combinations: Clarifying the Math

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SUMMARY

The discussion centers on calculating the number of possible license plate combinations consisting of three letters followed by three digits, with specific constraints on the letters and digits used. The correct calculation involves recognizing that the first letter is fixed (A), and the digits must include both 1 and 2 in any order, leading to a total of 40,560 combinations. Participants clarified the use of permutations, specifically P(3,3), to account for the arrangement of the digits and emphasized the importance of understanding the context of license plates, which may exclude certain letters like I, O, and Q.

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  • Understanding of basic combinatorial principles
  • Familiarity with permutations and the formula P(n, r) = n! / (n - r)!
  • Knowledge of the constraints in license plate formats
  • Basic arithmetic skills for calculating combinations
NEXT STEPS
  • Study combinatorial mathematics, focusing on permutations and combinations
  • Research the specific rules and formats for license plates in different jurisdictions
  • Learn about the application of combinatorial principles in real-world scenarios
  • Explore advanced counting techniques, including the inclusion-exclusion principle
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Mathematicians, educators, students studying combinatorics, and anyone interested in the practical applications of mathematical principles in real-world scenarios such as license plate generation.

  • #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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