Permutations, combinations and variations of negative numbers.

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SUMMARY

The discussion centers on the mathematical concepts of permutations, combinations, and variations involving negative numbers. Participants agree that using negative values, such as -2 in combinations, leads to nonsensical results based on the standard combinatorial formula. Specifically, the formula for combinations, \(\left( \begin{array}{c}n\\ m \end{array} \right) = \frac{n!}{m! \, (n-m)!}\), does not yield valid outcomes when negative integers are substituted for \(m\). The consensus is that the result of such combinations is defined as zero.

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  • Understanding of combinatorial mathematics
  • Familiarity with factorial notation and operations
  • Basic knowledge of LaTeX for mathematical expressions
  • Concept of permutations and combinations
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  • Research the implications of negative integers in combinatorial formulas
  • Study advanced combinatorial identities and their applications
  • Explore the concept of generating functions in combinatorics
  • Learn about the combinatorial interpretations of negative values
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Mathematicians, students studying combinatorics, educators teaching mathematical concepts, and anyone interested in the theoretical implications of negative numbers in combinatorial contexts.

sutupidmath
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need help on this?

well guys i was doing some problems with series and i cam up with this problem, i think it belongs to combinatorics but i'll post it here.
How is defined the permutations, combinations and variatons of negative numbers. For example if you were required to find the combinations of -2 elements taken from a set of n elements. TO me this makes no sens?

[tex]\left(\begin{array}{cc}-2\\&n)[/tex]

any help would be appreciated

p.s. sorry for my latex, but i do not know how to write this.
 
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Zero. Or not

sutupidmath said:
How is defined the permutations, combinations and variatons of negative numbers. For example ...[itex]\left( {\small \begin{array}{c}n \\ -2 \end{array} } \right)[/itex]

The standard answer is zero, but if you simply plug a negative value for m (or a value exceeding n) into the formula
[tex] \left( \begin{array}{c}n\\ m \end{array} \right) = \frac{n!}{m! \, (n-m)!}<br /> = \frac{n \, (n-1) \dots (n-m+1)}{ m \, (m-1) \, (m-2) \dots 1}[/tex]
you get nonsense. See this and this for some musings which should intrigue you!

sutupidmath said:
p.s. sorry for my latex, but i do not know how to write this.

Try hitting "reply" and look at the LaTeX markup in the window.
 
Last edited:

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