Chapter 12.5, Problem 46AYU

Show that if $n$ and $j$ arc integers with $0\le j\le n$, then, $\left(\begin{array}{l}n\\ j\end{array}\right)=\left(\begin{array}{l}n\\ n-j\end{array}\right)$ Conclude that the Pascal triangle is symmetric with respect to a vertical line drawn from the topmost entry.

To determine

To prove: To show that $n$ and $j$ are integers with $0\le j\le n$, then $\left(\begin{array}{c}n\\ j\end{array}\right)=\left(\begin{array}{c}n\\ n-j\end{array}\right)$.

Answer to Problem 46AYU

The last really simple thing to observe is that Pascal’s triangle is symmetric about the central axis. If we draw a vertical line down the page from the 1 in the 0 row, what appears on the left side of the triangle is the same as what appears on the right hand side of the triangle. This is because of the relation $\left(\begin{array}{c}n\\ j\end{array}\right)=\left(\begin{array}{c}n\\ n-j\end{array}\right)$.

Explanation of Solution

Pascal’s triangle is a sequence of rows that form a triangle that gets wider as we go down. Let’s label the rows as 0, 1, 2, . . . etc. For example:

$\begin{array}{llllllllllll}0\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \left(\begin{array}{c}0\\ 0\end{array}\right)\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ 1\hfill & \hfill & \hfill & \hfill & \hfill & \left(\begin{array}{c}1\\ 0\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}1\\ 1\end{array}\right)\hfill & \hfill & \hfill & \hfill & \hfill \\ 2\hfill & \hfill & \hfill & \hfill & \left(\begin{array}{c}2\\ 0\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}2\\ 1\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}2\\ 2\end{array}\right)\hfill & \hfill & \hfill & \hfill \\ 3\hfill & \hfill & \hfill & \left(\begin{array}{c}3\\ 0\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}3\\ 1\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}3\\ 2\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}3\\ 3\end{array}\right)\hfill & \hfill & \hfill \\ 4\hfill & \hfill & \left(\begin{array}{c}4\\ 0\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}4\\ 1\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}4\\ 2\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}4\\ 3\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}4\\ 4\end{array}\right)\hfill & \hfill \\ 5\hfill & \left(\begin{array}{c}5\\ 0\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}5\\ 1\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}5\\ 2\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}5\\ 3\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}5\\ 4\end{array}\right)\hfill & \hfill & \left(\begin{array}{c}5\\ 5\end{array}\right)\hfill \end{array}$

Here 0 row is formed by taking the element $\left(\begin{array}{c}0\\ 0\end{array}\right)$. The 1 row is listing the elements $\left(\begin{array}{c}1\\ 0\end{array}\right)$ and $\left(\begin{array}{c}1\\ 1\end{array}\right)$. The 2 row is listing the elements $\left(\begin{array}{c}2\\ 0\end{array}\right)$, $\left(\begin{array}{c}2\\ 1\end{array}\right)$ and $\left(\begin{array}{c}2\\ 2\end{array}\right)$ and so on.

The n row becomes $\left(\begin{array}{c}n\\ 0\end{array}\right),\left(\begin{array}{c}n\\ 1\end{array}\right),\left(\begin{array}{c}n\\ 2\end{array}\right),\dots ,\left(\begin{array}{c}n\\ n\end{array}\right)$. Further position the elements $\left(\begin{array}{c}n\\ j\end{array}\right)$ and $\left(\begin{array}{c}n\\ j\end{array}\right)$ so that the element $\left(\begin{array}{c}n\\ j\end{array}\right)$ is between them and one level above them.

As it stands, it is quite cumbersome to compute the rows of Pascal’s triangle. For example, to compute the 8 row, we have to compute each element from $\left(\begin{array}{c}8\\ 0\end{array}\right),\left(\begin{array}{c}8\\ 1\end{array}\right),\left(\begin{array}{c}8\\ 2\end{array}\right),\dots ,\left(\begin{array}{c}8\\ 8\end{array}\right)$. Instead of doing such hard work, let’s look at the table with the numbers $\left(\begin{array}{c}n\\ j\end{array}\right)$ computed up through the 8 row and look for patterns

$\begin{array}{llllllllllllllllll}n=0\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & 1\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ n=1\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & 1\hfill & \hfill & 1\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ n=2\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & 1\hfill & \hfill & 2\hfill & \hfill & 1\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ n=3\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & 1\hfill & \hfill & 3\hfill & \hfill & 3\hfill & \hfill & 1\hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ n=4\hfill & \hfill & \hfill & \hfill & \hfill & 1\hfill & \hfill & 4\hfill & \hfill & 6\hfill & \hfill & 4\hfill & \hfill & 1\hfill & \hfill & \hfill & \hfill & \hfill \\ n=5\hfill & \hfill & \hfill & \hfill & 1\hfill & \hfill & 5\hfill & \hfill & 10\hfill & \hfill & 10\hfill & \hfill & 5\hfill & \hfill & 1\hfill & \hfill & \hfill & \hfill \\ n=6\hfill & \hfill & \hfill & 1\hfill & \hfill & 6\hfill & \hfill & 15\hfill & \hfill & 20\hfill & \hfill & 15\hfill & \hfill & 6\hfill & \hfill & 1\hfill & \hfill & \hfill \\ n=7\hfill & \hfill & 1\hfill & \hfill & 7\hfill & \hfill & 21\hfill & \hfill & 35\hfill & \hfill & 35\hfill & \hfill & 21\hfill & \hfill & 7\hfill & \hfill & 1\hfill & \hfill \\ n=8\hfill & 1\hfill & \hfill & 8\hfill & \hfill & 28\hfill & \hfill & 56\hfill & \hfill & 70\hfill & \hfill & 56\hfill & \hfill & 28\hfill & \hfill & 8\hfill & \hfill & 1\hfill \end{array}$

Note that there are a few easy things to observe.

First, the entries on the left and right edges are alwalys 1. This is easy to explain, because they are always of the form $\left(\begin{array}{c}n\\ 0\end{array}\right)$ or $\left(\begin{array}{c}n\\ n\end{array}\right)$ and $\left(\begin{array}{c}n\\ 0\end{array}\right)=\left(\begin{array}{c}n\\ n\end{array}\right)=1$

Further, the entries next to the edges in each row just give the number of the row. For example, in the 6 row, the entries next to the 1’s on the outer edge are 6 and 6. This is because they represent $\left(\begin{array}{c}6\\ 1\end{array}\right)=6$ and $\left(\begin{array}{c}6\\ 5\end{array}\right)=6$. More generally, in the $n$ row, these entries are $\left(\begin{array}{c}n\\ 1\end{array}\right)=n$ and $\left(\begin{array}{c}n\\ n-1\end{array}\right)=n$, so this is why we always get $n$ for those entries.

The last really simple thing to observe is that Pascal’s triangle is symmetric about the central axis. If we draw a vertical line down the page from the 1 in the 0 row, what appears on the left side of the triangle is the same as what appears on the right hand side of the triangle. This is because of the relation $\left(\begin{array}{c}n\\ j\end{array}\right)=\left(\begin{array}{c}n\\ n-j\end{array}\right)$.