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Here is a direct screenshot of the book:enter image description here

First of all, what does type mean? Does the author mean that the set with $r$ elements can be partitioned into $n$ subsets? Secondly, an $r$ permutation of $n$ elements was previously defined as the number of ways to arrange $r$ out of $n$ elements in a row. How is that different than the permutation that is defined above? The title of the chapter is "Arrangements and Selections with repetitions", so I guess this has something to do with repetitions.

Side note: If the word "type" is not a strictly defined term which I am simply not aware of, what is the benefit of teaching combinatorics with vague terms like this instead of sticking to plain set notation/terminology?

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You are right that this has to do with repetitions. Here, "type" means that the objects of the same type are interchangeable. For instance, if you were looking at rearrangements of the word "ANAGRAM", then the A's would be indistinguishable, so instead of $7!$ arrangements you would have $7!/3!$ (where $r_1 = 3$ and $r_i = 1$ for $i \ne 1$).

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  • $\begingroup$ Well, if I can rearrange elements and it wouldnt make a difference, how is it different than arranging $n$ elements in a row, each representing a different type.Then the formula would be $n!$. I hope you understand my question. $\endgroup$
    – alexgiorev
    Apr 24 '15 at 20:45
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    $\begingroup$ You get the formula $n!$ when all of the elements are different types (corresponding to $r_1 = r_2 = \ldots = r_n = 1$, but if some of your elements are not distinguishable, $n!$ will count the same thing repeatedly. For instance, there are only 3 arrangements of the letters AAB, namely AAB, ABA, and BAA, where here the formula is $3!/2!$. If every letter was distinct you would have $6 = 3!$ instead, but that's not the case here. $\endgroup$
    – Rolf Hoyer
    Apr 24 '15 at 20:51
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For example, take colors to imagine : type $1$ is red, type $2$ is blue, and so on. So, for exmaple, imagine that you have $5$ balls, red $2$, blue $2$, white $1$, but each ball in the same color cannot be distinguished. I hope this helps.

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