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Section 5.5 Counting Techniques

Example 5.5.2. Committee Selection.

Three members from a 14 member committee are to be randomly selected to serve as chair, vice chair, and secretary. The first person selected is the chair, the second is the vice chair, and the third is the secretary. How many different committee structures are possible?

Solution
\begin{gather*} 14 \cdot 13 \cdot 12 = 2184 \end{gather*}
Example 5.5.3. Security Codes.

Outside a garage, there is a 10 digit keypad numbered 0 through 9. The correct 4 digit code will open the garage door. The numbers on the keypad can be repeated. How many codes are possible? What is the probability of getting the correct code on the first try assuming the person doesn’t know the code?

Solution
\begin{gather*} 10 \cdot 10 \cdot 10 \cdot 10 = 10^4 = 10,000 \end{gather*}

Thus a person has a one in 10,000 chance of guessing the correct code.

Example 5.5.4. Outfits.

A gentleman has 7 shirts, 3 vests and 5 suits. Assuming they all match, how many different combinations can he wear?

Solution
\begin{gather*} 7 \cdot 3 \cdot 5 = 105 \end{gather*}

Reading Questions Reading Questions

Find the value of each factorial.

1.

\(4!\)

Answer

\(4! = 4\cdot 3\cdot 2\cdot 1 = 24\)

2.

\(12!\)

Answer

\(12! = 12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 = 479,001,600\)

Example 5.5.6. How Many Routes?

You have just been hired as a book representative for Pearson Education. On your first day, you must travel to five schools to introduce yourself. How many different routes are possible?

Solution

\(5! = 5\cdot 4\cdot 3\cdot 2\cdot 1 = 120\)

Example 5.5.7. Scrabble Tile Combinations.

While playing Scrabble you have the letter tiles: A, E, I, S, D, P, L, R. How many tile combinations can you create?

Solution

\(8! = 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 = 40,320\)

Definition 5.5.8.

A permutation of objects is an ordered arrangement of the objects.

  1. The \(n\) objects are distinct (all different).
  2. Repetition of objects is not allowed.
  3. Order is important.
\begin{equation*} nPr = \frac{n!}{(n-r)!} \end{equation*}
Definition 5.5.9.

A combination is a collection, without regard to order, of objects.

  1. The \(n\) objects are distinct.
  2. Repetition of objects is not allowed.
  3. Order is not important.
\begin{equation*} nCr = \frac{n!}{(n-r)!r!} \end{equation*}
Example 5.5.10.

How many ways can different students from a class of 10 be assigned to work the first three homework problems on the board? Does order matter?

Solution

Order does matter, so we must compute the number of permutations.

\begin{equation*} 10 P 3 = \frac{10!}{(10-3)!} = \frac{10!}{7!} = \frac{10\cdot 9\cdot 8\cdot 7!}{7!} = 10\cdot 9\cdot 8 = 720 \end{equation*}
Example 5.5.11. Focus Group.

How many ways can a focus group of 5 consumers be selected from a list of 25 people who purchased a particular product? Notice that order doesn’t matter since we are just forming a focus group, so changing the order in which people are selected does not change the group.

Solution
\begin{align*} 25 C 5 =\amp \frac{25!}{(25-5)!5!}\\ =\amp \frac{25!}{20!5!}\\ =\amp \frac{25\cdot 24\cdot 23\cdot 22\cdot 21\cdot 20!}{20!5!}\\ =\amp \frac{25\cdot 24\cdot 23\cdot 22\cdot 21} {5\cdot 4\cdot 3\cdot 2\cdot 1}\\ =\amp 53,130 \end{align*}
Example 5.5.12. Volleyball Teams.

A group of four friends (Alan, Becky, Curtis and Dylan) is going to play volleyball by dividing up into two teams. How many different combinations of the 4 friends taken 2 at a time are there? List all the combinations.

Solution

Notice that order does not matter, and further notice that by picking one team, the other team is also determined.

\begin{equation*} 4 C 2 = \frac{4!}{(4-2)!\,2!} = \frac{4!}{2!\,2!} = \frac{4\cdot 3\cdot 2!}{2!\,2!} = 6 \end{equation*}

If we use the first letter of each name to represent each person, then the possible pairings are:

  1. \(\{(A,B), (C,D)\}\)
  2. \(\{(A,C), (B,D)\}\)
  3. \(\{(A,D), (B,C)\}\)
  4. \(\{(B,C), (A,D)\}\)
  5. \(\{(B,D), (A,C)\}\)
  6. \(\{(C,D), (A,B)\}\)

However, the last three pairings are repeats of the first three, so there are exactly 3 possible team pairs. Because choosing a team of two also forces the pairing of the other team, we get exactly half of \(4C2\text{.}\)

Example 5.5.13. Speed Skaters.

Suppose 24 speed skaters are competing in the Olympics. How many ways can the top 3 racers finish the race?

Solution
\begin{equation*} 24 P 3 = \frac{24!}{(24-3)!} = \frac{24!}{21!} = \frac{24\cdot 23\cdot 22\cdot 21!}{21!} = 12,144 \end{equation*}
Definition 5.5.14. Permutations with Non Distinct Items.
\begin{equation*} \frac{n!}{n_1! \, n_2!\, \cdots \, n_k!} \end{equation*}
  1. The \(n\) objects are not distinct, with \(k\) different types of objects.
  2. \(n_1\) is the number of the first type, \(n_2\) the number of the second type, etc.
  3. Repetition of objects is not allowed.
  4. Order is important.
Example 5.5.15. Flag Arrangements.

How many different vertical arrangements are there of 10 flags if 5 are white, 3 are blue, and 2 are red?

Solution
\begin{equation*} \frac{10!}{5!\, 3!\, 2!} = \frac{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5!} {5!\, (3\cdot 2\cdot 1)(2\cdot 1)} = 2,520 \end{equation*}
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