Expand Factorials: (n-2)! & (n-n')!

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    Expansion Factorial
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Discussion Overview

The discussion focuses on the expansion of factorial expressions, specifically (n-2)! and (n-n')!. Participants explore how to express these factorials in terms of their product forms, with an emphasis on understanding the notation and the process of expansion.

Discussion Character

  • Technical explanation, Conceptual clarification

Main Points Raised

  • One participant inquires about how to expand (n-n')!, questioning whether it simplifies to just (n-n').
  • Another participant clarifies that "expanding" a factorial means expressing it as a product of terms, suggesting the use of ellipses for longer products.
  • A further response provides an example of how to write n! as a product and implies that (n-n')! can be similarly expressed.
  • It is proposed that (n-n')! can be expanded to (n-n')(n-n'-1)(n-n'-2)...(2)(1), substituting (n-n') for n in the factorial definition.
  • One participant expresses gratitude for the clarification, indicating a level of understanding achieved.

Areas of Agreement / Disagreement

The discussion does not appear to have significant disagreement, as participants generally agree on the method of expanding factorials, though the initial inquiry reflects some uncertainty about the terminology.

Contextual Notes

There may be assumptions regarding the definitions of factorial and the notation used, which could affect the clarity of the discussion. The scope of the factorial expansion is limited to the expressions presented without exploring broader implications or applications.

khary23
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I know how to expand a factorial of the form (n-2)!. How would one expand (n-n')! ?
Would it just be (n-n')?
thanks in advance
 
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What do you mean expand ? The factorial is just a <shortening> of a product, so by my records <expanding> it means expressing the whole terms explicitely (if many, maybe use the <...> sign).
 
What I meant is how does one write (n-n')! For example n!= n(n-1)(n-2)...
 
n!=(n)(n-1)(n-2)...(2)(1)

Therefore, (n-n')! is just n-n' substituted for all n above, thus

(n-n')(n-n'-1)(n-n'-2)...(2)(1).
 
Ok thought it was that simple. thank you!
 

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