Binomial Coefficient

Binomial Coefficient

Binomial Coefficient

$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$$ Note that the binomial coefficient is sometimes written $^nC_k$ or $C^n_k$, depending on authors or geographical regions.

\( \begin{aligned} \binom{n}{k} &= \dfrac{n!}{k!(n-k)!} \cdots (1)\\ \binom{n}{n-k} &= \dfrac{n!}{(n-k)!(n-(n-k))!} = \dfrac{n!}{(n-k)!k!} \cdots (2)\\ \therefore \binom{n}{k} &= \binom{n}{n-k} \text{by } (1) \text{ and } (2) \\ \end{aligned} \)

This means;
\( \begin{aligned} \binom{10}{2} &= \binom{10}{8} \\ \binom{100}{1} &= \binom{100}{99} \\ \end{aligned} \)

The following binomial coefficients are undefined.

\( \displaystyle \binom{5}{7} = \dfrac{5!}{7! \times (5-7)!}\) $\rightarrow$ $(-2)!$ is undefined

$ \displaystyle \binom{5}{3.3} = \dfrac{5!}{3.3! \times (5-3.3)!}$ $\rightarrow$ $3.3!$ is undefined

$ \displaystyle \binom{5.2}{3} = \dfrac{5.2!}{5.2! \times (5.2-3)!}$ $\rightarrow$ $5.2!$ is undefined

$ \displaystyle\binom{5}{-2} = \dfrac{5!}{(-2)! \times (5-(-2))!}$ $\rightarrow$ $(-2)!$ is undefined

Relationship with Pascal's triangle

\begin{matrix} n=0&&&&&&1 \\ &&&&&&\binom{0}{0} \\ n=1&&&&&1&&1 \\ &&&&& \binom{1}{0}&& \binom{1}{1} \\ n=2&&&&1&&2&&1 \\ &&&& \binom{2}{0} && \binom{2}{1} && \binom{2}{2} \\ n=3&&&1&&3&&3&&1 \\ &&& \binom{3}{0} && \binom{3}{1} && \binom{3}{2} && \binom{3}{3} \\ n=4&&1&&4&&6&&4&&1 \\ && \binom{4}{0} && \binom{4}{1} && \binom{4}{2} && \binom{4}{3} && \binom{4}{4} \\ \end{matrix}

Example 1

Simplify $ \displaystyle \binom{n}{n-1}$.

Example 2

Evaluate $ \displaystyle \binom{8}{2}$.

Example 3

Evaluate $ \displaystyle \binom{8}{0}$.

Example 4

Evaluate $ \displaystyle \binom{4}{5}$.

Related YouTube video: Pascal’s Triangle using Combination