Chapter 12.5, Problem 45AYU

Show that $\left(\begin{array}{l}n\\ n-1\end{array}\right)=n\text{and}\left(\begin{array}{l}n\\ n\end{array}\right)=1$.

To determine

To prove: To show that $\left(\begin{array}{c}n\\ n - 1\end{array}\right) = n$ and $\left(\begin{array}{c}n\\ n\end{array}\right) = 1$.

Answer to Problem 45AYU

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.

Explanation of Solution

Given:

It is asked to prove that $\left(\begin{array}{c}n\\ n - 1\end{array}\right) = n$ and $\left(\begin{array}{c}n\\ n\end{array}\right)=1$.

Formula used:

The symbol $\left(\begin{array}{c}n\\ j\end{array}\right)$ is defined as

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

$\left(\begin{array}{c}n\\ n - 1\end{array}\right) = \frac{n!}{j!\left(n - j\right)!} = \frac{n!}{\left(n - 1\right)!\left(n - \left(n - 1\right)\right)!} = \frac{n!}{1! \times \left(n - 1\right)!} = n$

Therefore, $\left(\begin{array}{c}n\\ n - 1\end{array}\right) = n$

$\left(\begin{array}{c}n\\ n\end{array}\right) = \frac{n!}{j!\left(n - j\right)!} = \frac{n!}{\left(n\right)!\left(n - n\right)!} = \frac{n!}{0! \times n!} = 1$

Therefore, $\left(\begin{array}{c}n\\ n\end{array}\right) = 1$.

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

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 $\left(\begin{array}{c}n\\ 0\end{array}\right),\left(\begin{array}{c}n\\ 1\end{array}\right),\left(\begin{array}{c}n\\ 2\end{array}\right),\cdots ,\left(\begin{array}{c}n\\ n\end{array}\right)$. 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 $\left(\begin{array}{c}8\\ 0\end{array}\right),\left(\begin{array}{c}8\\ 1\end{array}\right),\left(\begin{array}{c}8\\ 2\end{array}\right),\cdots ,\left(\begin{array}{c}8\\ 8\end{array}\right)$. 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

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 $\left(\begin{array}{c}n\\ \mathrm{n}\end{array}\right)$ 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.