Number of ways to arrange letters

[evinda]

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)

 

[I like Serena]

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)

 

[evinda]

Yep.
It is right! (Mmm)

Nice,thank you! :)

 

[DavidCampen]

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.

 

[blue_raver22]

Hello! Your calculation is correct. This is known as a permutation problem, where the order of the letters matters. In this case, we have a total of 12 letters, with 5 A's, 3 B's, and 4 C's. To find the number of ways to arrange them, we use the formula nPr = n! / (n-r)!, where n is the total number of objects and r is the number of objects we are choosing. So in this case, we have 12 letters and we are choosing all 12 of them, giving us 12P12 = 12! / (12-12)! = 12! / 0! = 12!. However, since we have repeating letters, we need to divide by the factorial of the number of repeating letters to avoid counting the same arrangement multiple times. So we divide by 5! for the A's, 3! for the B's, and 4! for the C's, giving us a total of 12! / (5! * 3! * 4!) = 12! / (5! * 3! * 4!) = 12! / 5! * 3! * 4! = 12! / (5! * 3! * 4!). Therefore, your calculation is correct. Great job!