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$\begingroup$

I know I have to use the combination formula since the professor wrote it like this:

There are $3-O's$ and $2-W's$ then $\frac{9!}{(3!)(2!)}= 30240$

My question is the reason they have $(3!)(2!)$ as the denominator is because they are just repeated characters?

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    Yes, that's the reason. https://en.wikipedia.org/wiki/Permutation#Permutations_of_multisets – 2017-02-09
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    Oh okay, I was just making sure, thanks! – 2017-02-09

2 Answers 2

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Yes. We have total 9 characters and out of which O is repeating thrice and W is twice.

So in denominator 3! and 2!.

In general if we have to find in how many ways all letters of word are arranged.

Then we use,

$\frac{\text{(Number of total letters)!}}{\text{(No. of times 1 letter repeating)!} × \text{(No. of times other repeating)!} × ..}$

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    Thank you for the formula, I was just doubtful. – 2017-02-09
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    Mine pleasure :-) – 2017-02-09
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We have $(3!)\;\text{and}\;(2!)$ in denominator since we have repeated terms.

Suppose you says that the answer is $9!$ but in all these you include the cases in which you have same words.

Let us say the $\text{O's}$ be $O_1,O_2,O_3$ then the words may form be $BRO_1WNWO_2O_3D$ and $BRO_2WNWO_1O_3D$ and many others but as you can differ them here you cannot differ in between $\text{BROWNWOOD}$ and $\text{BROWNWOOD}$ since the $O's$ are not numbered so you cannot realise that I changed the position of $O's$. And as you take $9!$ as answer you count both of them.

So you need to divide by $3!\;\text{and}\;2!$

Hope it helps!!!