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I came across the text given below from the book Combinatorics - a problem oriented approach:

                            Binomial Expansions 

If we expand the expression $(A + B)^4$ by first writing $\texttt (A + B)(A + B)(A + B)(A + B)$,

We find that each term in the product comes from selecting either A or B from each of the four factors and multiplying them. The result is:

$\texttt (AAAA + AAAB + AABA + ABAA + BAAA + AABB + . . . + BBBB)= \\=A^4 + 4A^3 B + 6A^2 B^2 + 4AB^3 + B^4.$

Notice that the coefficient of $A^3 B$ represents the number of rearrangements of three $As$ and one $B$. In the next term, the coefficient of $A^2 B^2$ is the number of rearrangements of two $As$ and two $Bs$. This is $4\choose 2$= 6, since the two As can occupy any two of the four positions.

In general, the coefficient of $ A^k B^{\text (n-k)}$ in the expansion of $(A + B)^n$ is $n \choose k$ . So now we can write $\sum_{k=0}^{n}$ $n\choose k$ $A^k$ $B^{\text (n-k)}.$

This is the binomial expansion formula.

I'm in doubt after reading this that how is it rearrangement of $As$ and $Bs$ .Doesn't it look like permuting the $As$ and $Bs$. Please let me know why am I wrong.

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  • $\begingroup$ A "permutation" is usually used for distinct objects. I think that is why he is using the term rearrangement. He want to make clear that swapping the position of two As does not count as creating a different arrangement. $\endgroup$ – almagest Sep 8 '14 at 7:45
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The word "rearrangement" is used here probably to avoid confusion with either "permutation" or "combination", which are terms the reader is likely to have seen used, both with a slightly different meaning. The technically correct (but little used) term is "multiset permutation". It describes the following situation: while playing Scrabble, you are holding a bunch of letters, not necessarily distinct; the set of different "words" that you can make using all your letters is the set of multiset permutations of your collection of letters (the "multi" refers to the possible coincidences among these letters). Those words can be formed by "rearranging" the letters in the usual sense of that verb; "permuting" is just a synonym of "rearranging".

In the example the situation is particularly simple, as there are only two distinct letters $A,B$ involved. In that case each multiset permutation is determined by the positions chosen to have the letter $A$, and such a choice of positions is what is commonly referred to as a combination. But that correspondence is apparently not emphasised in the text (and it would break down if there were more than two distinct letters), so the more informal (and general) term "rearrangement of letters" is used.

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Here are the $\binom{4}{2} = 6$ combinations (in lexicographic order). $$ AABB \\ ABAB \\ ABBA \\ BAAB \\ BABA \\ BBAA \\ $$

In general, $\binom{n}{k}$ enumerates the number of words in the alphabet $\{A, B\}$ of length $n$, with $k$ many $A$s and $n-k$ many $B$s.

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  • $\begingroup$ What has this got to do with the question? $\endgroup$ – almagest Sep 8 '14 at 7:33
  • $\begingroup$ I actually can't find a specific question from the OP. I'm just trying to shine some light on why we count combinations rather than permutations. $\endgroup$ – Sammy Black Sep 8 '14 at 7:36
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    $\begingroup$ It seems to be a question about terminology. Why "rearrangement", rather than "permutation". $\endgroup$ – almagest Sep 8 '14 at 7:47

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