Talk:Petrolsoft Corporation

Product combinations

The article says that a tank truck with 4 compartments with 4 possible products can be filled 4⁴ ways. In practical terms it isn't important which part of the truck the fuel is in so the actual number of permutations will be less. - Stillwaterising (talk) 01:37, 1 April 2010 (UTC) If I remember high school math it would four factorial 4x3x2x1=24 unique combinations of fuels. - Stillwaterising (talk) 23:26, 1 April 2010 (UTC)

The same product can fill multiple compartments. Each of the four compartments can be filled with one of the four products. So there are four options for each of the four compartments yielding 4⁴ ways. If products could not fill multiple compartments, and each product had to fill exactly one compartment, then your four factorial calculation would be correct; however, this is not the case. I am a math teacher. Mathteacher69 (talk) 04:42, 2 April 2010 (UTC)

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Considering that cross-contamination is not an issue, then the compartments can be washout and be used for any type of fuel right?

Here's a simple example with 2 compartments (called A and B), with 3 fuels (D,G,& P). To show these visually:

Group#1 (same in both)

1. AG/BG
2. AD/BD
3. AP/BP

Group#2 (Gas in one, Premium in other)

1. AG/BP
2. AP/BG

Group#3 (Gas in one, Diesel in other)

1. AG/BD
2. AD/BG

Group#4 (Premium in one, Diesel in other)

1. AP/BD
2. BP/AD

Each combination in Group 1 is unique, so all 3 are counted. Groups 2-4 show combinations that deliver the same amount and types of fuel to the destination, with only 1 unique combinations of each. This makes for 6 unique combinations out of 9 permutations.

This example assumes that both compartments are of equal size. I'm not sure how the mathematics are worked out, but I'm sure it uses the laws of Permutations and Combinations. -Stillwaterising (talk) 10:53, 9 April 2010 (UTC)

I looked up the math, the solution would be combinations with repetition. The formula for that would be ${\frac {(n+r+1)!}{r!(n-1)!}}$ from this page here where n is the number of fuels, and r is the number of compartments.

To solve for the simple example above, (3+2+1)!/2!(3-1)!=4!/4=4/(4*3*2*1)=6 which works out the same as the visual example above.

Now to take this to the 4 compartments, and four fuel example which you correctly calculated the permutations as 4x4x4x4=256

The combinations are (4+4-1)!/(4!(4-1)!)=7!/(4! x 3!)=5040/(24 x 6) = 35 unique combinations of fuel. - Stillwaterising (talk) 10:50, 9 April 2010 (UTC)

• This is an interesting discussion, but I am afraid you are both still completely wrong about this due to the fact that both combination and permutation problems assume choosing among distinct items when you cannot choose the same item more than once. In determining the four compartment truck example (and compartments are usually different sizes to increase flexibility), you may choose the same product more than once. It is a Decision Chart problem using the Fundamental Counting Principle[1]. There are four decisions, one for each compartment (which of the four products are to go there). Each decision has four choices. To determine the number of ways all four decisions can be made together, you would multiply the number of ways each decision can be made together. In this case that is 4*4*4*4=256.Mathteacher69 (talk) 01:25, 13 April 2010 (UTC)

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