$$\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$$
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
\( \begin{matrix}
n=0&&&&&&1 \\
&&&&&&\displaystyle\binom{0}{0} \\
n=1&&&&&1&&1 \\
&&&&&\displaystyle \binom{1}{0}&& \displaystyle\binom{1}{1} \\
n=2&&&&1&&2&&1 \\
&&&&\displaystyle \binom{2}{0} &&\displaystyle \binom{2}{1} &&\displaystyle \binom{2}{2} \\
n=3&&&1&&3&&3&&1 \\
&&&\displaystyle \binom{3}{0} &&\displaystyle \binom{3}{1} &&\displaystyle \binom{3}{2} &&\displaystyle \binom{3}{3} \\
n=4&&1&&4&&6&&4&&1 \\
&&\displaystyle \binom{4}{0} &&\displaystyle \binom{4}{1} &&\displaystyle \binom{4}{2} &&\displaystyle \binom{4}{3} &&\displaystyle \binom{4}{4}
\end{matrix} \)
Example 1
Simplify $ \displaystyle \binom{n}{n-1}$.
\( \begin{align} \displaystyle
\binom{n}{n-1} &= \dfrac{n!}{(n-1)! \times 1!} \\
&= \dfrac{n \times (n-1)!}{(n-1)!} \\
&= n
\end{align} \)
Example 2
Evaluate $ \displaystyle \binom{8}{2}$.
\( \begin{align} \displaystyle
\binom{8}{2} &= \dfrac{8!}{2! \times 6!} \\
&= \dfrac{8 \times 7 \times 6!}{2 \times 1 \times 6!} \\
&= \dfrac{8 \times 7}{2} \\
&= 28
\end{align} \)
Example 3
Evaluate $ \displaystyle \binom{8}{0}$.
\( \begin{align} \displaystyle
\binom{8}{0} &= \dfrac{8!}{0! \times 8!} \\
&= \dfrac{8!}{1 \times 8!} \\
&= 1
\end{align} \)
Example 4
Evaluate $ \displaystyle \binom{4}{5}$.
\( \begin{align} \displaystyle
\binom{4}{5} &= \dfrac{4!}{5! \times (4-5)!} \\
&= \dfrac{4!}{5! \times (-1)!} \\
&= \text{undefined}
\end{align} \)
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