Pascal’s Triangle and the Binomial Theorem

On this page we discuss an important algebra theorem which helps expand arbitrary large integer powers of a sum, the so-called Binomial Theorem. In passing, we also discuss its relationship to Pascal’s Triangle and the so-called Binomial Coefficients which are important in the field of Combinatorics (and therefore in Probability & Statistics).

Pascal’s Triangle and Binomial Coefficients

Pascal’s Triangle is the following infinite triangle of numbers, where the first and last numbers in each row are always \(1\), and all other numbers are obtained by taking the sum of the two numbers directly above in the previous row:

\begin{gathered} 1\\ 1 \quad 1\\ 1 \quad 2 \quad 1\\ 1 \quad 3\quad 3\quad 1\\ 1 \quad 4\quad 6\quad 4\quad 1\\ 1\quad 5\quad 10\quad 10\quad 5\quad 1\\ 1\quad 6\quad 15\quad 20\quad 15\quad 6\quad 1\\ \vdots \end{gathered}

These numbers have many different applications and show up in surprising places, but they are most important in the field of Combinatorics, which deals with the theory of counting various combinations of things. From that context, we introduce a notation for each of the numbers in the triangle: The number in row \(n\) and (diagonal) column \(k\) (both starting at \(0\)) is denoted by \(\binom{n}{k}\) and is read as “\(n\) choose \(k\)” and also called a “Binomial Coefficient”:

\begin{gathered} \binom{0}{0}\\ \binom{1}{0} \quad \binom{1}{1}\\ \binom{2}{0}\quad \binom{2}{1} \quad \binom{2}{2}\\ \binom{3}{0} \quad \binom{3}{1}\quad \binom{3}{2}\quad \binom{3}{3}\\ \binom{4}{0} \quad \binom{4}{1}\quad \binom{4}{2}\quad \binom{4}{3}\quad \binom{4}{4}\\ \binom{5}{0} \quad \binom{5}{1}\quad \binom{5}{2}\quad \binom{5}{3}\quad\binom{5}{4}\quad \binom{5}{5}\\ \binom{6}{0} \quad \binom{6}{1}\quad \binom{6}{2}\quad\binom{6}{3}\quad\binom{6}{4}\quad \binom{6}{5}\quad \binom{6}{6}\\ \vdots \end{gathered}

For small \(n\) and \(k\), it is easiest to just quickly write down the first few rows of Pascal’s Triangle to quickly find these numbers, but it turns out that there is also a direct formula:

\begin{equation} \boxed{\binom{n}{k}=\frac{n!}{k!\, (n-k)!}} \end{equation}

where \(n!=n\cdot(n-1)\cdot (n-2)\cdots 3\cdot 2\cdot 1\) is called “\(n\) factorial”, which is defined by \(0!=1\), \(1!=1\), \(n!=(n-1)!\cdot n\).

As an example, you should verify that

\begin{equation} \binom{5}{3}=\frac{5!}{3!\cdot (5-3)!}=\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1 \cdot (2\cdot 1)} = \frac{5\cdot 4}{2}=10 \end{equation}

Note that we can (and should) cancel common factors intelligently when evaluating this. Using this direct formula (and the cancellation) is in particular useful for large values of \(n\) and \(k\), for example:

\begin{align} \binom{2019}{2015}&=\frac{2019!}{2015!\cdot (2019-2015)!}=\frac{2019!}{2015!\cdot 4!}\\&= \frac{2019\cdot 2018\cdot 2017\cdot 2016}{4\cdot 3\cdot 2\cdot 1} = \frac{2019}{3}\cdot \frac{2018}{2}\cdot 2017\cdot \frac{2016}{4}\\&= 673\cdot 1009 \cdot 2017 \cdot 504 = \boxed{690\,307\,616\,376} \end{align}

Hint

Most scientific calculators have a button to directly calculate Binomial Coefficients, though they often use the notation \({}_n C_k\) instead of \(\binom{n}{k}\).

There are incredibly many interesting things one can discover about the numbers in Pascal’s Triangle, and new ones are still discovered today! This fun video showcases some of the more unusual properties of Pascal’s Triangle you might not have heard about yet:

The Binomial Theorem

By calculating the first few powers of a sum \((a+b)\) by expanding by hand, you should observe a surprisingly familiar pattern:

\begin{align} (a+b)^0&= 1\\ (a+b)^1&=a+b\\ (a+b)^2&=a^2+2ab+b^2\\ (a+b)^3&=a^3+3a^2b+3ab^2+b^3\\ (a+b)^4&=a^4+4a^3b+6a^2b^2+4ab^3+b^4\\ (a+b)^5&=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5 \end{align}

This is of course no coincidence, and it has to do with the way the terms are collected after the expansion. The number of each particular combination of powers of \(a\) and \(b\) is given precisely by a number from Pascal’s Triangle!

More precisely, we have:

Binomial Theorem: For any positive integer \(n\), we have:

\begin{align} (a+b)^n &= \sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k\\ &= \binom{n}{0}a^nb^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots+\binom{n}{n-1}a^1b^{n-1}+\binom{n}{n}a^0 b^n\\ &= a^n + na^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots+nab^{n-1}+ b^n \end{align}

Note in particular the symmetry, and the fact that the first two coefficients in the expansion for power \(n\) are \(1\) and \(n\).

Watch the following video which shows how to use this theorem in practice:

Calculus Application of the Binomial Theorem

In Calculus, we can use the Binomial Theorem to find the derivative of the function \(f(x)=x^n\) for any positive integer \(n\geq 2\), using the limit-definition of the derivative:

\begin{align} f'(x)&=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to 0} \frac{(x+h)^n-x^n}{h}\\ &=\lim_{h\to 0} \frac{{\color{red}{x^n}}+nx^{n-1}h+\binom{n}{2}x^{n-2}h^2+\cdots+n x h^{n-1}+h^n-{\color{red}{x^n}}}{h}\\ &=\lim_{h\to 0} \frac{nx^{n-1}{\color{blue}{h}}+\binom{n}{2}x^{n-2}{\color{blue}{h}}^2+{\color{blue}{\cdots}}+n x {\color{blue}{h}}^{n-1}+{\color{blue}{h}}^n}{h}\\ &=\lim_{h\to 0} \frac{{\color{blue}{h}}\left(nx^{n-1}+\binom{n}{2}x^{n-2}h^1+\cdots+n x h^{n-2}+h^{n-1}\right)}{h}\\ &\overset{h\neq 0}{=} \lim_{h\to 0}\quad nx^{n-1}+\binom{n}{2}x^{n-2}{\color{green}{h}}^1+{\color{green}{\cdots}}+n x {\color{green}{h}}^{n-2}+{\color{green}{h}}^{n-1}\\ &\overset{{\color{green}{h=0}}}{=} nx^{n-1}+{\color{green}{0}}+{\color{green}{\cdots}} + {\color{green}{0}}+{\color{green}{0}}\\ &=\boxed{nx^{n-1}} \end{align}

Pascal’s Triangle and the Binomial Theorem

Pascal’s Triangle and Binomial Coefficients

The Binomial Theorem

Calculus Application of the Binomial Theorem

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