Question

Show that if n and j are integers with $0\le j\le n,then\left(\begin{array}{c}n\\ j\end{array}\right)=\left(\begin{array}{c}n\\ n-j\end{array}\right)$

Conclude that the Pascal triangle is symmetric with respect

to a vertical line drawn from the topmost entry.

Short answer

If j and n are integers with$0\le j\le nthen\left(\begin{array}{c}n\\ j\end{array}\right)=\frac{n!}{j!(n-j)!}$

Step-by-step solution

  Step 1. Given information 

Here we have to show that$for0\le j\le n,then\left(\begin{array}{c}n\\ j\end{array}\right)=\left(\begin{array}{c}n\\ n-j\end{array}\right)$

Step 2. Showing  nj=nn-j

We know that

$$\begin{gathered} \left(\begin{array}{c}n\\ j\end{array}\right)=\frac{n!}{j!(n-j)!},jandnintegers0\le j\le n \\ then\left(\begin{array}{c}n\\ n-j\end{array}\right)=\frac{n!}{(n-n+j)!(n-j)!} \\ =\frac{n!}{j!(n-j)!} \\ =\left(\begin{array}{c}n\\ j\end{array}\right) \end{gathered}$$

Step 3. Drawing the Pascal triangle 

By using $\left(\begin{array}{c}n\\ j\end{array}\right)$,we can arrange values in a triangular form

$$\begin{gathered} \left(\begin{array}{c}0\\ 0\end{array}\right) \\ \left(\begin{array}{c}1\\ 0\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right) \\ \left(\begin{array}{c}2\\ 0\end{array}\right)\left(\begin{array}{c}2\\ 1\end{array}\right)\left(\begin{array}{c}2\\ 2\end{array}\right) \\ \left(\begin{array}{c}3\\ 0\end{array}\right)\left(\begin{array}{c}3\\ 1\end{array}\right)\left(\begin{array}{c}3\\ 2\end{array}\right)\left(\begin{array}{c}3\\ 3\end{array}\right) \end{gathered}$$

now we can draw Pascal triangle is symmetric with respect

to a vertical line drawn from the topmost entry.

$$\begin{gathered} n=0,j=01 \\ n=1,j=111 \\ n=2,j=2121 \\ n=3,j=31331 \end{gathered}$$