I recognize I need to use the mix formula due to the fact that the professor wrote it prefer this:

There are $3-O"s$ and also $2-W"s$ climate $\frac9!(3!)(2!)= 30240$

My concern is the factor they have actually $(3!)(2!)$ as the denominator is since they are just repeated characters?


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Yes. Us have full 9 characters and also out of i beg your pardon O is repeating thrice and W is twice.

So in denominator 3! and 2!.

You are watching: Find the number of distinguishable arrangements of each of the following "words."

In basic if we have to find in how numerous ways all letter of word are arranged.

Then we use,

$\frac\text(Number of complete letters)!\text(No. Of times 1 letter repeating)! × \text(No. Of times other repeating)! × ..$


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We have actually $(3!)\;\textand\;(2!)$ in denominator since we have actually repeated terms.

Suppose you claims that the prize is $9!$ but in all these you incorporate the instances in i m sorry you have same words.

Let us say the $\textO"s$ be $O_1,O_2,O_3$ climate the words may kind be $BRO_1WNWO_2O_3D$ and $BRO_2WNWO_1O_3D$ and many others yet as you can differ them here you can not differ in in between $\textBROWNWOOD$ and also $\textBROWNWOOD$ due to the fact that the $O"s$ are not numbered so you cannot realise the I changed the position of $O"s$. And as you take it $9!$ as answer you counting both that them.

So you must divide by $3!\;\textand\;2!$

Hope it helps!!!


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Harsh KumarHarsh Kumar
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