Binomial Coefficients & Pascal's Triangle (DP IB Analysis & Approaches (AA): HL): Revision Note

The binomial coefficient ⁿC_r

What is the formula for ${}_{\mathit{ }}{}^{\mathit{n}}{\mathbf{C}}_r$?

• The formula for is ${}_{ }{}^n{\mathrm{C}}_r$ is

${}_{ }{}^n{\mathrm{C}}_r= \frac{n!}{r!\left(n-r\right)!}$

• where the factorial symbol $!$ means, for example:

  • $4!=4\times 3\times 2\times 1=24$

• e.g. ${}_{}{}^{5}C_2=\frac{5!}{2!\left(5-2\right)!}=\frac{5!}{2!3!}=\frac{5\times 4\times \cancel{3\times 2\times 1}}{2\times 1\times \cancel{3\times 2\times 1}}=\frac{20}{2}=10$

What properties of ${}_{\mathit{ }}{}^{\mathit{n}}{\mathbf{C}}_r$ do I need to know?

• Useful properties of ${}_{ }{}^n{\mathrm{C}}_r$ to know are

  • $0\le r\le n$

  • ${}_{ }{}^n{\mathrm{C}}_0=1$

    • note that $0!=1$

  • ${}_{ }{}^n{\mathrm{C}}_n=1$

  • The numbers ${}_{ }{}^n{\mathrm{C}}_0,{}_{ }{}^n{\mathrm{C}}_1,{}_{ }{}^n{\mathrm{C}}_2,...,{}_{ }{}^n{\mathrm{C}}_{n-2},{}_{ }{}^n{\mathrm{C}}_{n-1},{}_{ }{}^n{\mathrm{C}}_n$ are symmetric

    • i.e. ${}_{ }{}^n{\mathrm{C}}_r={}_{ }{}^n{\mathrm{C}}_{n-r}$

How does ${}_{\mathit{ }}{}^{\mathit{n}}{\mathbf{C}}_{\mathit{r}}$ relate to the binomial theorem?

• The binomial theorem gives you the expansion of ${\left(a+b\right)}^n$ for different positive integer powers of $n$:

$\begin{array}{rcl}{\left(a+b\right)}^n& =& a^n+{}_{}{}^n{\mathrm{C}}_1{a}^{n-1}b+...+{}_{ }{}^n{\mathrm{C}}_r{a}^{n-r}{b}^r+...+b^n\\ {}_{ }{}^n{\mathrm{C}}_r& =& \frac{n!}{r!\left(n-r\right)!}\end{array}$

• where ${}_{}{}^{n}C_r$ is the binomial coefficient

How does ${}_{\mathit{ }}{}^{\mathit{n}}{\mathbf{C}}_r$ relate to counting principles?

• The value of ${}_{ }{}^n{\mathrm{C}}_r$ represents the number of ways to choose $r$ objects out of $n$ different objects

  • This is an example of a counting principle

  • e.g. how many ways can you choose 2 people out of 5 people?

    • There are ${}_{}{}^{5}C_2=\frac{5!}{2!\left(5-2\right)!}=10$ ways to do this

How is the binomial theorem related to counting principles?

• You can relate ${}_{ }{}^n{\mathrm{C}}_r$ in the binomial theorem to the number of ways to choose $r$ objects out of $n$ different objects

• ${\left(a+b\right)}^n$ means multiply $\left(a+b\right)$ by itself $n$ times

  • i.e. ${\left(a+b\right)}^n=\left(a+b\right)\left(a+b\right)...\left(a+b\right)$

  • Without using the binomial theorem, expand the right-hand side

    • i.e. multiplying all combinations of taking a letter from each bracket

• e.g. one possibility is $a\times a\times a...\times a\times b\times b$ where the last 2 are $b$'s

  • The first $\left(n-2\right)$ are $a$'s

  • This simplifies to $a^{n-2}{b}^2$

  • But you can also choose 2 $b$'s out of another 2 brackets (not just the last 2)

    • How many ways can you chose 2 $b$'s out of the $n$ brackets?

  • This is number of ways to choose $r$ objects out of $n$ different objects

    • i.e. there are ${}_{}{}^{n}C_2$ ways to do this

  • so the full term is ${}_{}{}^{n}C_2{a}^{n-2}{b}^2$

Pascal's triangle

What is Pascal’s triangle?

• Pascal’s triangle is formed as follows:

  • The first row and second row form a triangle of 1s

  • Every row below starts and ends with a 1

  • The middle terms are found by

    • adding the two terms above it

How does Pascal’s triangle relate to binomial coefficients ${}_{\mathit{ }}{}^{\mathit{n}}{\mathbf{C}}_r$?

• The binomial coefficients ${}_{ }{}^n{\mathrm{C}}_0,{}_{ }{}^n{\mathrm{C}}_1,{}_{ }{}^n{\mathrm{C}}_2,...,{}_{ }{}^n{\mathrm{C}}_{n-2},{}_{ }{}^n{\mathrm{C}}_{n-1},{}_{ }{}^n{\mathrm{C}}_n$ are rows in Pascal’s triangle

• e.g. calculate ${}_{}{}^{5}C_1,{}_{}{}^{5}C_2,{}_{}{}^{5}C_3,{}_{}{}^{5}C_4,{}_{}{}^{5}C_5$

  • If you use the formula $\begin{array}{rcl}{}_{ }{}^n{\mathrm{C}}_r& =& \frac{n!}{r!\left(n-r\right)!}\end{array}$

    • e.g. ${}_{}{}^{5}C_0=\frac{5!}{0!\left(5-0\right)!}=\frac{5!}{5!}=1$ (as $0!=1$)

    • e.g. ${}_{}{}^{5}C_1=\frac{5!}{1!\left(5-1\right)!}=\frac{5!}{4!}=\frac{5\times \cancel{4}\times \cancel{3}\times \cancel{2}\times \cancel{1}}{\cancel{4}\times \cancel{3}\times \cancel{2}\times \cancel{1}}=5$

    • etc

  • you get $1,5,10,10,5,1$

  • This is the same as the $1,5,10,10,5,1$ row in Pascal's triangle!

How does Pascal’s triangle relate to the binomial theorem?

• The binomial theorem gives you the expansion of ${\left(a+b\right)}^n$ for different positive integer powers of $n$:

$\begin{array}{rcl}{\left(a+b\right)}^n& =& a^n+{}_{}{}^n{\mathrm{C}}_1{a}^{n-1}b+...+{}_{ }{}^n{\mathrm{C}}_r{a}^{n-r}{b}^r+...+b^n\\ {}_{ }{}^n{\mathrm{C}}_r& =& \frac{n!}{r!\left(n-r\right)!}\end{array}$

• Instead of using $\begin{array}{rcl}{}_{ }{}^n{\mathrm{C}}_r& =& \frac{n!}{r!\left(n-r\right)!}\end{array}$ to calculate

  • ${}_{ }{}^n{\mathrm{C}}_0,{}_{ }{}^n{\mathrm{C}}_1,{}_{ }{}^n{\mathrm{C}}_2,...,{}_{ }{}^n{\mathrm{C}}_{n-2},{}_{ }{}^n{\mathrm{C}}_{n-1},{}_{ }{}^n{\mathrm{C}}_n$

  • just read the values off the relevant row in Pascal's triangle!