Combinatorics of license plates

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SUMMARY

The discussion focuses on combinatorial problems related to license plates and polygons. For the license plate identifier consisting of four digits and two letters (W, X, Y, Z), the total valid combinations are calculated as 32,000, derived from 2,000 four-digit combinations divisible by 5 multiplied by 16 letter combinations. In the context of a decagon, it is established that there are 35 diagonals when excluding sides. The discussion also touches on the equivalence of six-digit integers, with a calculation suggesting 848,800 non-equivalent integers, although further clarification is needed on the constraints involving digits 0 and 9.

PREREQUISITES
  • Understanding of combinatorial mathematics
  • Knowledge of divisibility rules, specifically for 5
  • Familiarity with polygon properties, particularly diagonals
  • Basic principles of permutations and equivalence classes in integers
NEXT STEPS
  • Explore combinatorial counting techniques in depth
  • Learn about the properties of polygons and their diagonals
  • Study permutations and combinations in number theory
  • Investigate advanced divisibility rules and their applications
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Mathematics students, educators, and anyone interested in combinatorial problems and their applications in real-world scenarios such as license plate generation and geometric properties.

Ryuuken
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Homework Statement



1. In the manufacture of commercial license plates, a valid identifier consists of four digits followed by two eltters. Among all possible plate identifiers how many contain only the letters W, X, Y, or Z with a four digit number divisible by 5?

2. All the vertices of a decagon are to be connected by straight lines called the diagonals.

a. If a side of a decagon does not count as a diagonal, then how many diagonals can be drawn?

b. If the decagon is drawn so that no more than two diagonals intersect at anyone point, then into how many line segments will the diagonals be divided by the intersecting diagonals?

3. Consider the set of 6-digit itnergers, where leading 0's are permitted. Two integers are considered to be "equivalent" if one can be obtained from the other by a permutation of the digits. Thus 129450 and 051294 are "equivalent". Among all the 10^6 six digit integers"

a. How many are non-equivalent integers are there?

b. If digits 0 and 9 can appear at most once, how many non-equivalent integers are there?

Homework Equations





The Attempt at a Solution



1. 10 * 10 * 10 * 10 = 10000 possibly number combinations

4 * 4 = 16 possible letter combinations

10000/5 = 2000 divisible by 5

16 * 2000 = 32000 combos <-- Is this correct?

2a. Each vertice can make 7 diagonals not including the sides. There are 10 sides so there are 70 diagonals. Since some vertices share the same diagonals, there are 70/2 = 35 diagonals.

2b. Is there a formula for this?

3a. Is it 10^6 - P(10, 6) = 848800?

3b. ...

Thanks.
 
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Ryuuken said:
4 * 4 = 16 possible letter combinations

I think you should have 26 * 26, no?

Edit: My mistake, 4 * 4 is right, I read too fast.
 

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