Chapter 5, Problem 1E

To determine

To prove: The Pascal's formula by replacing the values of the binomial coefficients as given in $\left(\begin{array}{l}n\\ k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}$.

Explanation of Solution

Formula used:

The pascal’s triangle formula is:

$\begin{array}{c}\left(\begin{array}{l}n\\ k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}\\ =\frac{n\left(n-1\right)\cdot \cdot \cdot \left(n-k+1\right)}{k\left(k-1\right)\cdot \cdot \cdot 1}\end{array}$

Calculation:

For an integer k and a real number n, to show the given formula.

$\left(\begin{array}{l}n\\ k\end{array}\right)=\left(\begin{array}{l}n-1\\ k-1\end{array}\right)+\left(\begin{array}{c}n-1\\ k\end{array}\right)\cdot \cdot \cdot \left(\text{i}\right)$

From equation $\left(\text{i}\right)$ first use the formula of left-hand side value:

$\begin{array}{c}\left(\begin{array}{l}n\\ k\end{array}\right)=\frac{P\left(n,k\right)}{k!}\\ =\frac{nP\left(n-1,k-1\right)}{k!}\end{array}$

And for the right-side of the equation $\left(\text{i}\right)$ value $\left(\begin{array}{l}n-1\\ k-1\end{array}\right)$ is:

$\begin{array}{c}\left(\begin{array}{l}n-1\\ k-1\end{array}\right)=\frac{P\left(n-1,k-1\right)}{\left(k-1\right)!}\\ =\frac{kP\left(n-1,k-1\right)}{k!}\end{array}$

And for the right-side of the equation $\left(\text{i}\right)$ value $\left(\begin{array}{c}n-1\\ k\end{array}\right)$ is:

$\begin{array}{c}\left(\begin{array}{c}n-1\\ k\end{array}\right)=\frac{P\left(n-1,k\right)}{k!}\\ =\frac{\left(n-k\right)P\left(n-1,k-1\right)}{k!}\end{array}$

So, put the value of the $\left(\begin{array}{l}n-1\\ k-1\end{array}\right)$ and $\left(\begin{array}{c}n-1\\ k\end{array}\right)$ in $\left(\begin{array}{l}n-1\\ k-1\end{array}\right)+\left(\begin{array}{c}n-1\\ k\end{array}\right)$.

$\begin{array}{c}\left(\begin{array}{l}n-1\\ k-1\end{array}\right)+\left(\begin{array}{c}n-1\\ k\end{array}\right)=\frac{kP\left(n-1,k-1\right)}{k!}+\frac{\left(n-k\right)P\left(n-1,k-1\right)}{k!}\\ =\frac{P\left(n-1,k-1\right)}{k!}\left(k+n-k\right)\\ =\frac{nP\left(n-1,k-1\right)}{k!}\end{array}$

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

Hence proved the Pascal's formula by replacing the values of the binomial coefficients.