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From my textbook:

The expansion of $(x + y)^3$ can be found using combinatorial reasoning instead of multiplying the three terms out. When $(x + y)^3 = (x + y)(x + y)(x + y)$ is expanded, all products of a term in the first sum, a term in the second sum, and a term in the third sum are added. Terms of the form $x^3$, $x^2y$, $xy^2$, and $y^3$ arise.

What does the bolded part mean?

I found that if you find the possible combinations of $x$ and $y$, you can get $xxx = x^3$, $xxy = x^2y$, $xyy = xy^2$, $yyy = y^3$. Is this what it means?

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  • $\begingroup$ It just means all products of a term from the first $x+y$, the second $x+y$ and the third $x+y$ are added. To get the coefficient on a term $x^ky^3-k$ you count the number of ways to choose $k$ $x$'s from the three $(x+y)$'s $\endgroup$
    – smcc
    Jul 24, 2016 at 23:12

1 Answer 1

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Expanding $(x+y)(x+y)(x+y)$ amounts to adding up all the ways you can pick three factors to multiply together. For example, you could pick an $x$ from the first $(x+y)$, a $y$ from the second $(x+y)$, and another $x$ from the third $(x+y)$ to get $xyx=x^2 y$.

You are right, the only possible products we can get are $x^3$, $x^2 y$, $xy^2$, and $y^3$. However, we do need to count how many ways to get each factor. For example there is only one way to get $x^3$ (pick $x$ from each $(x+y)$), but there are three ways to get $x^2 y$: $xxy$, $xyx$, and $yxx$. One way to count this is to realize that there are $3$ ways to choose which $(x+y)$ contributes a $y$ [and the rest will be $x$s]. Similar reasoning for $xy^2$ and $y^3$ shows that the expansion is $x^3 + 3x^2 y + 3 xy^2 + y^3$.

In general, if you have $(x+y)^n$, the number of ways to obtain a product of the form $x^k y^{n-k}$ is the number of ways to choose $k$ of the $(x+y)$ factors from which to select an $x$. There are $\binom{n}{k}$ ways to make this choice. This proves the binomial theorem $(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k n^{n-k}$.

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  • $\begingroup$ do you know a clean proof and formulation of this 'generalized distributive law' ?(of course it should follow from the usual distributive law) $\endgroup$
    – M. Van
    Jul 25, 2016 at 0:10
  • $\begingroup$ @M.Van The first thing that comes to mind is usual distributive law + induction, but this may not be as clean/elegant as what you are looking for... $\endgroup$
    – angryavian
    Jul 25, 2016 at 0:12
  • $\begingroup$ Thanks! I understood the later parts of the expansion but wasn't sure if I understood the beginning correctly! $\endgroup$ Jul 27, 2016 at 5:01

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