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A, B and C are prices of a good at $3$ points in time respectively. How many price combinations are possible. Please provide me a formula as well.

Example,

  1. $A=B=C$

  2. $A>B>C$

  3. $A>B$ and $C>A$

  4. $A>B$ and $C<A$

  5. $A>B$ and $A>C$ and $B>C$

  6. $A>B$ and $A<C$ and $B>C$

  7. $A>B$ and $A=C$ and $B>C$

so the possible combinations should consider "$>$" ,"$<$" "$ \lor $", " $=$ " in every possible way

Noa Even
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Teja
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3 Answers3

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With $2$ "$=$", there is $1$ chance. $\\$

With $1$ "$=$", there is $3+3$ chances. ($3!/2$ different ordering for first $>$ then $=$ (Ex: $a>b=c$ (note that $a>b=c$ and $a>c=b$ is same)) and $3!/2$ different ordering for first $=$ then $>$ (Ex: $a=b>c$ (note that $a=b>c$ and $b=a>c$ is same)) ) $\\$

With $0$ "$=$", there is $6$ chances. ($3!$ different ordering (Ex: $a>b>c$)) $\\$

Total $=13$ chances.

Taha Direk
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1

You have

or $$number <number < number$$

or $$number = number < number$$

or $$number < number = number$$

So for each string you have 6 posibilities in 1. case and 3 posibilities in case 2. and 3. that is 12 posibilities. You have also posibilite that all are equal so you have 13 posibilites in total.

What about $$number >number > number\;\;\;?$$ Well, all those are already counted in $number <number < number$. Say $a>b>c$ is the same as $c<b<a$.

nonuser
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    What about $a=b<c$ and $b=a<c$? You have counted those as two arrangements, but aren't they the same? (I suppose you could argue they are distinct, but then to be consistent you would have to count $a=b=c$ and $b=a=c$ as distinct also.) – awkward Jul 14 '19 at 12:37
  • Yes, you are right! @awkward Thanks for pointing that out – nonuser Jul 14 '19 at 12:38
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These are called "Fubini numbers" or the number of "preferential arrangements": http://oeis.org/A000670

RobPratt
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