Chapter 9.1, Problem 47E

Solution

To identify: The number which is not equal to 1 if $r\text{and}n$ are positive integers with $r<n$ .

The options are:

  $\left(a\right)\left(n-n\right)!$

  $\left(b\right)P_{n_n}$

  $\left(c\right)C_{n_n}$

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

  $\left(e\right)\left(\begin{array}{c}n\\ r\end{array}\right)÷\left(\begin{array}{c}n\\ n-r\end{array}\right)$

The correct option is $\left(b\right)P_{n_n}$ .

Given information:

It is given that $r\text{and}n$ are positive integers with $r<n$ .

Explanation:

Consider the option $\left(a\right)\left(n-n\right)!$ and find its value,

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

Consider the option $\left(b\right)P_{n_n}$ and find its value,

  $\begin{array}{c}{P}_{n_n}=\frac{n!}{\left(n-n\right)!}\\ =\frac{n!}{0!}\\ =n!\end{array}$

Option $\left(b\right)P_{n_n}$ is not equal to 1.

Now consider the option $\left(c\right)C_{n_n}$ and find its value,

  $\begin{array}{c}{C}_{n_n}=\frac{n!}{n!\left(n-n\right)!}\\ =\frac{n!}{n!0!}\\ =1\end{array}$

Consider the option $\left(d\right)\left(\begin{array}{c}n\\ n\end{array}\right)$ and find its value,

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

Consider the option $\left(e\right)\left(\begin{array}{c}n\\ r\end{array}\right)÷\left(\begin{array}{c}n\\ n-r\end{array}\right)$ and find its value,

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

Therefore, it can be seen that the answer of $P_{n_n}$ is not equal to 1. Option $\left(b\right)P_{n_n}$ should be the correct answer.