Chapter 10.5, Problem 17E

To determine

To calculate:

The different ways can the students finish first, second and third

Answer to Problem 17E

The different ways can the students finish first, second and third is $990$ .

Explanation of Solution

Given information:

  $\begin{array}{l}n=11\\ r=3\end{array}$

Calculation:

There are eleven students and out of eleven student’s three students to be selected, in order is very important.

The formula for permutation for $n$ objects taken $r$ at the time is as:

  $P{}_r=\frac{n!}{\left(n-r\right)!}\cdot \cdot \cdot \left(1\right)$

Where,

  $\begin{array}{l}n=11\\ r=3\end{array}$

Put the values in equation (1)

  $\begin{array}{l}P{}_3=\frac{11!}{\left(11-3\right)!}=\frac{11!}{8!}\\ P{}_3=\frac{11\times 10\times 9\times 8!}{8!}\\ P{}_3=990\end{array}$

The solution of the expression $P{}_3$ is $990$ . Hence, the different ways can the students finish first, second and third is $990$ .