Number of ways to arrange letters

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SUMMARY

The discussion centers on calculating the number of distinct arrangements of letters, specifically 5 letters A, 3 letters B, and 4 letters C. The correct formula for this calculation is given by the multinomial coefficient: $$\frac{(5+3+4)!}{5! \cdot 3! \cdot 4!}=\frac{12!}{5! \cdot 3! \cdot 4!}$$. Participants confirm the correctness of this approach and explore alternative expressions for the same calculation, emphasizing the importance of understanding the underlying combinatorial principles.

PREREQUISITES
  • Understanding of combinatorial mathematics
  • Familiarity with factorial notation
  • Knowledge of multinomial coefficients
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the concept of multinomial coefficients in combinatorics
  • Learn about permutations and combinations
  • Explore applications of combinatorial methods in probability theory
  • Practice solving problems involving arrangements of letters and objects
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Students studying combinatorics, educators teaching mathematical concepts, and anyone interested in solving problems related to arrangements and permutations.

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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