Chapter 12.6, Problem 19E

To determine

To find: Evaluation of $\left(\begin{array}{l}100\\ 98\end{array}\right)$ using pascal’s triangle.

Answer to Problem 19E

  $\Rightarrow \left(\begin{array}{l}100\\ 98\end{array}\right)=4950$

Explanation of Solution

Given information:

  $\left(\begin{array}{l}100\\ 98\end{array}\right)$

Calculation:

The formula for the expression $\left(\begin{array}{l}n\\ k\end{array}\right)$ is,

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

From given information, $n=8$ and $k=3$

  $\begin{array}{l}\Rightarrow \left(\begin{array}{l}100\\ 98\end{array}\right)=\frac{100!}{\left(100-98\right)!98!}\\ \Rightarrow \left(\begin{array}{l}100\\ 98\end{array}\right)=\frac{100\times 99\times 98\times \mathrm{.....}\times 4\times 3\times 2\times 1}{\left(2\times 1\right)\left(98\times 97\times 96\times \mathrm{......3}\times 2\times 1\right)}\\ \Rightarrow \left(\begin{array}{l}100\\ 98\end{array}\right)=\frac{100\times 99}{\left(2\times 1\right)}\\ \Rightarrow \left(\begin{array}{l}100\\ 98\end{array}\right)=\frac{9900}{2}\\ \Rightarrow \left(\begin{array}{l}100\\ 98\end{array}\right)=4950\end{array}$