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Section 2.6 The Binomial Theorem

Here is a truly basic result from combinatorics kindergarten.

View \((x+y)^n\) as a product

\begin{equation*} (x+y)^n=\underbrace{(x+y)(x+y)(x+y)(x+y)\dots(x+y)(x+y)}_{n\text{ factors} }. \end{equation*}

Each term of the expansion of the product results from choosing either \(x\) or \(y\) from one of these factors. If \(x\) is chosen \(n-i\) times and \(y\) is chosen \(i\) times, then the resulting product is \(x^{n-i}y^i\text{.}\) Clearly, the number of such terms is \(C(n,i)\text{,}\) i.e., out of the \(n\) factors, we choose the element \(y\) from \(i\) of them, while we take \(x\) in the remaining \(n-i\text{.}\)

Example 2.31.

There are times when we are interested not in the full expansion of a power of a binomial, but just the coefficient on one of the terms. The Binomial Theorem gives that the coefficient of \(x^5y^8\) in \((2x-3y)^{13}\) is \(\binom{13}{5}2^{5}(-3)^8\text{.}\)