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Exercises 8.5 Chapter Test

1.

In how many different ways can five people line up at the grocery store?

Answer.
120
Solution.
\begin{equation*} 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120 \end{equation*}
2.

How many different 4-digit pin numbers are possible?

Answer.
10,000
Solution.
\begin{equation*} 10 \cdot 10 \cdot 10 \cdot 10 = 10,000 \end{equation*}
3.

A computer password is made up of 4 characters. Each character can be a capital letter (A through Z), a lowercase letter (a through z), or a digit (0 through 9).

  1. How many different computer passwords are possible?

  2. How many passwords have no repeated characters?

  3. How many passwords start with a letter?

  4. How many passwords have 2 letters and 2 digits?

Answer.

  1. 14,776,336

  2. 13,388,280

  3. 12,393,056

  4. 1,622,400

Solution.

  1. There are \(26 + 26 + 10 = 62\) characters possible for each position in the passoword.

    \begin{equation*} 62 \cdot 62 \cdot 62 \cdot 62 = 14,776,336 \end{equation*}

  2. \begin{equation*} 62 \cdot 61 \cdot 60 \cdot 59 = 13,388,280 \end{equation*}

  3. \begin{equation*} 52 \cdot 62 \cdot 62 \cdot 62 = 12,393,056 \end{equation*}

  4. There are 6 different ways a password can have two digits (D) and two letters (L):

    L-L-D-D, L-D-L-D, L-D-D-L, D-L-D-L, D-L-L-D, D-D-L-L

    Each of these possiblities can occur in \(52 \cdot 52 \cdot 10 \cdot 10 = 270,400\) ways.

    There are a total of \(6 \cdot 270,000 = 1,622,400\) ways we can have two letters and two digits.

4.

Compute the following:

  1. \(\displaystyle _{8}P_{3}\)

  2. \(\displaystyle _{9}C_{4}\)

Answer.

  1. 336

  2. 126

Solution.

  1. \begin{equation*} _nP_r = \frac{n!}{(n-r)!} \end{equation*}
    \begin{equation*} _8P_3 = \frac{8!}{5!}=\frac{8\cdot 7 \cdot 6 \cdot 5!}{5!}=8\cdot 7 \cdot 6 =336 \end{equation*}

  2. \begin{equation*} _nC_r = \frac{n!}{(n-r)!r!} \end{equation*}
    \begin{equation*} _9C_4 = \frac{9!}{5!4!}=\frac{9\cdot 8 \cdot 7 \cdot 6 \cdot 5!}{5!4!} \end{equation*}
    \begin{equation*} _9C_4 = \frac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1}= \frac{3024}{24}=126 \end{equation*}

Exercise Group.
Use permutations or combinations to find the following.
5.

Three employees are selected from a group of 25 to attend a workshop. In how many ways can this occur?

Answer.
\(_{25}C_{3}=2,300\)
Solution.

Order doesn't matter, so use combinations.

\begin{equation*} _{25}C_{3}=\frac{25!}{22!3!}=\frac{25 \cdot 24 \cdot 23}{3 \cdot 2 \cdot 1}= 2300 \end{equation*}
6.

20 drivers start a race. Assuming all drivers are equally skilled, how many top five finishes are possible?

Answer.
\(_{20}P_{5}=1,860,480\)
Solution.

Order does matter, so use permutations.

\begin{equation*} _{20}P_{5}=\frac{20!}{15!}=20 \cdot 19 \cdot 18 \cdot 17 \cdot 16 = 1,860,480 \end{equation*}
7.

In how many ways can you select four books from your library of 30 books to bring on vacation?

Answer.
\(_{30}C_{4}=27,405\)
Solution.

Order doesn't matter, so use combinations.

\begin{equation*} _{30}C_{4}=\frac{30!}{26!4!}=\frac{30 \cdot 29 \cdot 28 \cdot 27}{4 \cdot 3 \cdot 2 \cdot 1}= 27,405 \end{equation*}