Number of ways to arrange letters

  • Context: MHB 
  • Thread starter Thread starter evinda
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Discussion Overview

The discussion revolves around the problem of determining the number of ways to arrange a set of letters, specifically 5 letters A, 3 letters B, and 4 letters C. The scope includes mathematical reasoning and combinatorial calculations.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Debate/contested

Main Points Raised

  • One participant proposes the formula for arrangements as $$\frac{(5+3+4)!}{5! \cdot 3! \cdot 4!}$$ and seeks confirmation of its correctness.
  • Another participant agrees with the initial calculation, stating it is correct.
  • A different approach is suggested by another participant, using combinations to arrive at the same result, but with a different expression involving binomial coefficients.
  • This participant expresses confusion regarding the educational approach of the instructor or textbook, questioning the lack of methods or proofs provided alongside the problem.

Areas of Agreement / Disagreement

There is some agreement on the correctness of the initial calculation, but a different method is introduced that suggests alternative perspectives on the problem. The discussion remains somewhat unresolved regarding the educational context and the reasoning behind the problem's presentation.

Contextual Notes

Participants express uncertainty about the educational purpose of the problem, highlighting a lack of methods or proofs in the instructional material.

evinda
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Hey again! :)

I am given this exercise:
With how many ways can we arrange $5$ letters $A$, $3$ letters $B$ and $4$ letters $C$?

I thought that it is :

$$\frac{(5+3+4)!}{5! \cdot 3! \cdot 4!}=\frac{12!}{5! \cdot 3! \cdot 4!}$$

Could you tell me if it is right? (Blush)
 
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evinda said:
Hey again! :)

I am given this exercise:
With how many ways can we arrange $5$ letters $A$, $3$ letters $B$ and $4$ letters $C$?

I thought that it is :

$$\frac{(5+3+4)!}{5! \cdot 3! \cdot 4!}=\frac{12!}{5! \cdot 3! \cdot 4!}$$

Could you tell me if it is right? (Blush)

Yep.
It is right! (Mmm)
 
I like Serena said:
Yep.
It is right! (Mmm)

Nice,thank you! :)
 
I would have said that the answer is:
\[(\begin{matrix}12\\5\end{matrix})(\begin{matrix}7\\4\end{matrix})=(\begin{matrix}12\\4\end{matrix})(\begin{matrix}8\\5\end{matrix})=
(\begin{matrix}12\\3\end{matrix})(\begin{matrix}9\\4\end{matrix}) etc.=\frac{12!}{3!\cdot4!\cdot5!}\]

I don't understand the purpose of Evinda's instructor or textbook asking for so many answers to be computed only without also supplying methods or proofs.
 

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