Calculating n!(N-n)!: N = 16, n = 3

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Discussion Overview

The discussion revolves around calculating the expression for combinations, specifically the formula for \( \frac{N!}{n!(N-n)!} \) with \( N = 16 \) and \( n = 3 \). Participants explore how to arrive at the numerical result of 560.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the formula \( \frac{16!}{3!13!} \) and expresses uncertainty about how to compute the result of 560.
  • Another participant suggests calculating \( \frac{16 \times 15 \times 14}{3 \times 2 \times 1} \) as a method to find the answer.
  • A third participant explains that \( 16! \) can be expressed as \( 16 \times 15 \times 14 \times 13! \), indicating that the \( 13! \) cancels out, which aligns with the previous suggestion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method of calculation, but there is agreement on the approach of simplifying the factorial expression. The discussion remains focused on the calculation process without resolving the initial participant's uncertainty.

Contextual Notes

The discussion does not address potential limitations or assumptions in the factorial calculations, nor does it clarify the reasoning behind the numerical result of 560.

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N!
____
n!(N - n)!

N = 16 n = 3

so,

16!
____
3!13! = 560

i don't know how to get the 560 answer?
 
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(16*15*14)/(3*2*1)
 
You can write a factorial like 16! as 16*15*14*13!.

The 13! then cancels on top and bottom, leaving what gnome indicated.

- Warren
 
thanks
 

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