Question

Question: Let r and n be integers with 0 < r < n. Prove that$\left[\begin{array}{c}n\\ r\end{array}\right]\mathbf{=}\left[\begin{array}{c}n\\ n-r\end{array}\right]$.

Short answer

Answer:

It is proved that $\left[\begin{array}{c}\mathrm{n}\\ \mathrm{r}\end{array}\right]=\left[\begin{array}{c}\mathrm{n}\\ \mathrm{n}-\mathrm{r}\end{array}\right]$.

Step-by-step solution

Binomial Coefficient

Foe each r, with 0 < r < n, the binomial coefficient $\left[\begin{array}{c}n\\ r\end{array}\right]$is defined to be the number $\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\left(n-r\right)\mathbf{!}}$.

Step 2: Prove that  [nr]=[nn-r].

Let r and n be integers with 0 < r < n.

For the right hand side equation, apply the binomial coefficient to $\left[\begin{array}{c}n\\ n-r\end{array}\right]$.

$$\begin{gathered} \left[\begin{array}{c}n\\ n-r\end{array}\right]=\frac{n!}{\left(n-r\right)!\left(n-\left(n-r\right)\right)!} \\ =\frac{n!}{\left(n-r\right)!r!} \end{gathered}$$

For the left hand side equation, apply the binomial coefficient to $\left[\begin{array}{c}n\\ r\end{array}\right]$.

$\left[\begin{array}{c}n\\ r\end{array}\right]=\frac{n!}{r!\left(n-r\right)!}$

Hence, it is proved that$\left[\begin{array}{c}n\\ r\end{array}\right]=\left[\begin{array}{c}n\\ n-r\end{array}\right]$.