Binomial theorem

Let's try to expand \((x+y)^4\):

\[(x+y)^4 = (x+y)(x+y)(x+y)(x+y) \]

Expanding:

\[(x+y)^4 = xxxx + \begin{aligned} &xxxy + \\ &xxyx + \\ &xyxx + \\ &yxxx \end{aligned} + \begin{aligned} &xxyy + \\ &xyxy + \\ &yxxy + \\ &xyyx + \\ &yxyx + \\ &yyxx \end{aligned} + \begin{aligned} &yyyx + \\ &yyxy + \\ &yxyy + \\ &xyyy \end{aligned} + yyyy \]

The first group contain 4 \(x\)'s, the second group contains 3 \(x\)'s and 1 \(y\)'s, the third group contains 2 \(x\)'s and 2 \(y\)'s, the fourth group contains 1 \(x\)'s and 3 \(y\)'s, and the fifth group contains 4 \(y\)'s. Let's look at the third group, with 2 \(x\)'s and 2 \(y\)'s.

According to this article, we know that there are \(\binom n k\) ways to arranging \(n\) items with \(k\) belonging to one group and the others belonging to another group. For the second group above, there are a total of \(\binom 4 1\) ways to arrange 4 items, with 1 belonging to one group (\(y\)), and the rest belonging to another group (\(x\)). For the third group above, there are a total of \(\binom 4 2\) ways to arrange 4 items, with 2 belonging to one group (\(y\)) and 2 belonging to the other group (\(x\)). A similar argument can be made for every other group.

\[(x+y)^4 = \binom 4 0 x^4 + \binom 4 1 x^3y^1 + \binom 4 2 x^2y^2 + \binom 4 3 x^1 y^3 + \binom 4 4 y^4 \]

We can genelarize this and say:

\[\begin{align} (x+y)^n &= \binom n 0 x^n + \binom n 1 x^{n-1} y^1 + \binom n 2 x^{n-2} y^2 + \cdots \\ &= \sum^n_{i=0} \binom n i x^{n-i} y^i \end{align}\]

Binomial Theorem

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