a simple combinatorics (i assume) problem

Nov 2014
2
0
Bulgaria
I feel that kind of problem should be very easy, but still got no clue how to solve it.(Thinking) Cars have 4 wheels, trikes - 3 wheels, trucks - 6 wheels. What's the number of all possible vehicle combinations that have a total amount of wheels of 50. Thanks.
 

Plato

MHF Helper
Aug 2006
22,492
8,653
I feel that kind of problem should be very easy, but still got no clue how to solve it. Cars have 4 wheels, trikes - 3 wheels, trucks - 6 wheels. What's the number of all possible vehicle combinations that have a total amount of wheels of 50.
If one expands $\left( {\sum\limits_{k = 0}^{13} {{x^{4k}}} } \right) \cdot \left( {\sum\limits_{k = 0}^{17} {{x^{3k}}} } \right) \cdot \left( {\sum\limits_{k = 0}^9 {{x^{6k}}} } \right)$
done here the term $20x^{50}$ is in the expansion. That answer your question.

As anyone can see it is far from a simple question. I have no idea to give a complete unit on generating functions in this sort of setting. But that is what is required if you really need to understand how it is solved.

That is: there are twenty ways to solve $4A+3B+6C=50$ in the non-negative integers.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
I think it is a relatively simple problem, though tedious! No, that is not combinatorics, this is a "Diophantine" equation, an equation to be solved for integer values. Let "x" be the number of cars, y the number or trikes, and z the number of trucks. Since "Cars have 4 wheels, trikes - 3 wheels, trucks - 6 wheels" the total number of wheels is 4x+ 3y+ 6z= 50. Find all combinations, (x, y, z) with x, y, z non-negative integers. The first thing I would notice is that since 6*9= 54> 50, there cannot be more than 8 trucks. With z trucks then the number of cars and trikes would satisfy 4x+ 3y= 50- 6z.

Since 8 is not all that large, you could now go over the 8 cases:
If z= 0, 4x+ 3y= 50
4- 3= 1 so 4(50)+ 3(-50)= 50 and x= 50, y= -50 is one solution. Of course, here, y= -50 is not acceptable- both x and y must be non-negative. But it is clear that x= 50- 3k, y= -50+ 4k is also a solution for any integer k: 4(50- 3k)+ 3(-50+ 4k)= 200- 12k- 150+ 12k= 50.
In order that y be non-negative, k must be at least 13. In order that x be non-negative, k cannot be larger than 16.
If k= 13, x= 50- 39= 11, y= -50+ 52= 2. x= 11, y= 2, z= 0 is one solution.
If k= 14, x= 50- 42= 8, y= -50+ 56= 6. x= 8, y= 6, z= 0 is another solution.
If k= 15, x= 50- 45= 5, y= -50+ 60= 10. x= 5, y= 10, z= 0 is another solution.
If k= 16, x= 50- 48 = 2, y= -50+ 64= 14. x= 2, y= 14, z= 0 is another solution.

If z= 1, 4x+ 3y= 44
Again, since 4- 3= 1, 4(44)+ 3(-44)= 44 so one solution is x= 44, y= -44. Also as before, x= 44- 3k, y= -44+ 4k. In order that y be non-negative, k must be at least 11 and in order that x be non-negative, k must be no more than 14.
If k= 11, x= 44- 33= 11,y= -44+ 44= 0. x= 11, y= 0, z= 1 is another solution
etc.

If z= 2, 4x+ 3y= 38
If z= 3, 4x+ 3y= 32
If z= 4, 4x+ 3y= 26
If z= 5, 4x+ 3y= 20
If z= 6, 4x+ 3y= 14
If z= 7, 4x+ 3y= 8
If z= 8, 4x+ 3y= 2
 
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