Chapter 8.2, Problem 8.37E

(a)

To determine

To compute the value of $P$ and explain why this value make sense.

Answer to Problem 8.37E

  $1$

Explanation of Solution

Use the definition of permutation:

  $\begin{array}{l}P=\frac{6!}{(6-0)!}\\ =\frac{6!}{6!}\\ =1\end{array}$

This makes sense because there is one way to pick zero people from six people that is to pick no one.

(b)

To determine

To compute the value of $P$ and $C$ also create an example involving six friends to show these values are related.

Answer to Problem 8.37E

  $\begin{array}{l}P=120\\ C=20\end{array}$

Explanation of Solution

Use the definition of permutation:

  $\begin{array}{l}P=\frac{6!}{(6-3)!}\\ =6\times 5\times 4\\ =120\end{array}$

Use the definition of combination:

  $\begin{array}{l}C=\frac{6!}{(6-3)!3!}\\ =\frac{6\times 5\times 4}{3\times 2\times 1}\\ =20\end{array}$

For example, seating three of the six friends around a round table can be done in $20$ ways while you can place them in $120$ ways next to each other at a rectangular table.

(c)

To determine

To calculate $C$ and $C$ also create an example involving six friends to show these values are related.

Answer to Problem 8.37E

  $\begin{array}{l}C=15\\ C=15\end{array}$

Explanation of Solution

Use the definition of permutation:

  $\begin{array}{l}C=\frac{6!}{(6-2)!2!}\\ =\frac{6\times 5}{2\times 1}\\ =15\end{array}$

Use the definition of combination:

  $\begin{array}{l}C=\frac{6!}{(6-4)!4!}\\ =\frac{6\times 5}{2\times 1}\\ =15\end{array}$

For example, pick four of the six friends to be on the debate team gives the same number of ways to pick them as picking two of the six friends to be on the team.