Online Explanation 7.4

In this section we will be studying three concepts, permutations, combinations, and the multiplication principle of the last section applied to permutations and combinations. Permutations and combination refer to the arrangement of objects in a set. It is helpful to keep one image of each in your head.

For permutation think of a 50-50 raffle at a sporting event. If 464 tickets were sold and the winning number was 412. The holder of 412 gets \$232 dollars. The holders of 124, 142, 214, 241, and 421 get nothing. What is important with permutation is the order the objects, in this case digits, are arranged in is important.

Permutation (Tan 443)

For combination think of a 5 card closed poker game. Say you were dealt the ace of hearts, ace of clubs, ace of diamonds, 7 of spades, and 7 of hearts, you have a full house, no matter what order you arrange them in your hand. Selecting r objects, 5 cards, from a set of n objects, 52 cards, without any regard to order means we are dealing with a combination problem.

Combination (Tan 449)

Permutations

A permutation of the set is an arrangement of the elements of the sets in which the order of the elements is taken into consideration.

Example 1 Permutations of Three Objects

Let A = {a, b, c}

a. Find the number of permutations of A.

b. List all the permutations with the aid of a tree diagram.

Each permutation consists of a sequence of three letters, taken from the set elements a, b, and c. Think of all possible ways to create the sequence. We could think of this as 3 blanks.

Then we would have any one of the three letters to fill in blank 1, any one of the two remaining letters to fill in blank 2, and the remaining letter to fill in blank 3.

Using the multiplication principle, we conclude that there are 3 times 2 times 1, or 6 permutations of set A.

The tree diagram above is a way to list all six permutations of A, abc, acb, bac, bca, cab, cba.

Problem 1 Student ID - Permutations

"Each student at Provincial University has a student I.D. number consisting of 7 digits (the first digit is nonzero, and digits can be repeated) followed by two of the letters A, B, C and D (letters cannot be repeated). How many different student numbers are possible?" (BCA Finite Math).

 n-Factorial
(Tan 445)

For example,

Note: To use our calculator there are three basic ways of using the probability operators of n!, nPr(n,r), and nCr(n,r). If we are repeatedly using the same operator, displaying it in the catalog area and use the catalog selector to enter it in the edit line works nicely. If we are switching between various combination of the probability operators, we can use the "math" button, with "option 7: Probability", to select the operation we need. Or if the operator appears in the pretty print area, we can use the "blue arrows" and "enter" to transfer the old problem to the edit area where it can be modified.

When we permuting a set of n distinct objects taken n at a time. Then n-factorial formula will get us the answer. This would be the case with our example 1, but not with problem 1.

We now present a formula for n distinct objects taken r at a time.

 Permutations of n Distinct Objects
(Tan 446)

Example 2 Permutation of n Taken r at a Time

Compute P(4,4) and interpret your results.

Since P(n,n) = n!, P(4,4) = 4! = 24. For our four object taken 4 at a time there are 24 arrangements or pemutations.

Problem 2 Expressions Involving Factorials

Example 3 Permutation of n Taken r at a Time

Let A = {a, b, c, d}

Compute the number of permutations of set A taken two at a time. Use fill in blanks and the multiplication principle or the formula

Display the above permutations with the aid of a tree diagram.

Problem 3 Permutation of n Taken r at a Time

How many three-digit numbers can be formed using the numerals in the set {3, 2, 7, 9} if repetition is not allowed?

Example 4 Committee - Officers - Permutation

"Find the number of ways a chairman, a vice-chairman, a secretary, and a treasurer can be chosen from a committee of eight members" (Tan 447).

Here we can again fill in four blanks starting with 8 members of the committee, then 7, 6, 5 for the jobs of chairperson, a vice-chairperson, secretary and treasurer and use the multiplication principle. The other approach would be to use P(8,4) = 1680. The final approach, the tree diagram, would be too time consuming.

Problem 4 Permutation of n Taken r at a Time

Howmany three-letter permutations can be formed from the first five letters of the alphabet?

 Permutations of n Objects, Not All Distinct

Example 5 Permutations of n Objects, Not all Distinct

"Weaver and Kline, a stock brokerage firm, has received nine inquiries regarding new accounts. In how many ways can these inquiries be directed to three of the firms account executives if each account executive is to handle three inquiries?" (Tan 448).

We use the tan's analogy to slots and business cards. Each account executive's business cards are alike, but different from the other two account exective's (Tan 448). Thus we would want to use n = 9, n1 = n2 = n3 = 3 and not all distinct permutation rule.

There would be 1680 ways of assigning the inquires.

Problem 5 Permutations of n Objects, Not all Distinct

"Find the number of distinguishable permutations that can be formed from the letters of the word PHILIPPINES" (Tan 456)

Combinations

We now turn to the situation when selecting r object from a set of n object where the order in which the objects are selected is unimportant. A subset of this type is called a combination.

 Combinations of n Distinct Objects
(Tan 449)

Example 6 Combination of n Taken r at a Time

Compute C(4,2) and interpret your results.

The number of combination of four distinct objects taken two at a time is 6.

Problem 6 Combination of n Taken r at a Time

Evaluate C(12,7)

Applications

Example 7 Committee Selection - Combination

"A Senate investigation subcommittee of four members is to be selected from a Senate committee of ten members. Determine the number of ways this can be done" (Tan 450).

Unlike the previous committee problem where we were designating officiers and order was important because it meant which office the person had. In this problem the order for members of the subcommittee is not important. To find the ways that this can be done we calculate C(10,4).

C(10,4) = 210

Problem 7 Combination of n Taken r at a Time

"Wayne allows Marcel to take three hockey cards at random from his collection of 181 different cards in exchange for a Gaston Gingras rookie card. How many different sets of three cards can be selected by Marcel?" (BCA Finite Math).

Example 8 Combination of n Taken r at a Time

"A 6/44 lottery requires choosing six of the numbers 1 through 44. Repeat numbers are not allowed. How many different lottery tickets can you choose?" (BCA Finite Math).

Suppose the winning numbers were 11, 14, 25, 34, 37, 44. The question is would 44, 34, 11, 14, 37, 25 and other tickets in different order also be winners? Since the order is not important in determining the winner in a lottery, we need to solve the problem C(44,6) to find how many different lottery tickets you can choose.

C(44,6) = 7059052

Problem 8 Combination of n Taken r at a Time

"A 6/47 lottery requires choosing six of the numbers 1 through 47. Repeat numbers are not allowed. How many different lottery tickets can you choose?" (BCA Finite Math).

Example 9 Poker - Combination of n Taken r at a Time

How many poker hands of 5 cards can be dealt from a standard deck of 52 cards?

The order that the dealer dealt the cards is not important in analyzing your hand. For example, it did not matter if you got 3 aces in the beginning, middle, end, or mixed inbetween. You still have a nice hand. The number of poker hands would be C(52,5), or 2598960.

Problem 9 Combination - Multiplication Principle

How many five-card poker hands can be dealt consisting of three queens and a pair of aces?

Example 10 Combination - Permutation - Multiplication Principle

"The members of a string quartet composed of two violinists, a violist, and a cellist are to be selected from a group of six violinists, three violists, and two cellists, respectively.

a. In how many ways can the string quartet be formed?

b. In how many ways can the string quartet be formed if one of the violinists

is to be designated as the first violinist and the other is to be designated as the second violinist?" (Tan 451).

In the first part of this question, we would use C(6,2), or 15 ways for the violinists; C(3,1), or 3 ways for the violist; C(2,1), or 2 ways for the cellist. Applying the multiplication principle we have

of forming the string quartet.

In the second part of the question, we would use P(6,2), or 30 ways for violinist where one is designated as the first violinist; the rest would be the same as above. Applying the multiplication principle we have

of forming the string quartet with a first violinist and a second violinist.

Note: In the above example second part we used both permutation and combination with the multiplication principle.

Problem 10 Permutation with an Entry being a Permutation

"A company car that has a seating capacity of six is to be used by six employees who have formed a car pool. If only four of these employees can drive, how many possible seating arrangements are there for the group?" (Tan 456).

Example 11 Investments - Combination - Multiplication Principle

"Suppose the investor has decided to purchase shares in the stocks of two aerospace companies, two energy development companies, and two electronics companies. In how many ways may the investor select the group of six companies for the investment from the recommended list of five aerospace companies, three energy development companies, and four electronics companies?

The answer is C(5,2)C(3,2)C(4,2), or 180 ways of selecting the six companies for her investment" (Tan 451).

Problem 11 Jury Selection - Combination - Multiplication Principle

"In how many different ways can a panel of 12 jurors and 2 alternate jurors be chosen from a group of 30 prospective jurors?" (Tan 456).

Example 12 Scheduling - Permutations - Multiplication Principle

"The Futurists, a rock group, are planning a concert tour with performances to be given in five cities: San Francisco, Los Angeles, San Diego, Denver, and Las Vegas. In how many ways can they arrange their itinerary if:

a. There are no restrictions?

b. The three performances in California must be given consecutively?" (Tan 451-452).

Arranging their itinerary with no restriction they can perform their concert tour in P(5,5), or 120 ways. Here the order of the itinerary would be important.

The more logical approach is once is California, do all the California stops consecutively. Now what we have is a two part problem. The first part is that we have three stops, Denver, Las Vegas, and California. The second part is once we are in California, we have three stops, San Francisco, Los Angeles, and San Diego.

Both part one and part two are P(3,3), or 6 ways. Using the multiplication principle we have 6 times 6, or 36 ways of arranging their itinerary.

Problem 12 Committees - Combination - Multiplication Principle

"In how many ways can a subcommittee of four be chosen from a Senate committee of five Democrats and four Republicans if:

a. All members are eligible?

b. The subcommittee must consist of two Republicans and two Democrats?" (Tan 457).

Example 13 U.N. - Combination - Multiplication - Addition Principles

"The U.N. Security Council consists of 5 permanent members and 10 nonpermanent members. Decisions made by the council require nine votes for passage. However, any permanent member may veto a measure and thus block its passage. In how many ways can a measure be passed if all 15 members of the Council vote (no abstentions)? (Tan 452).

We must use the multiplication principle and multiply the permament voting for by the nonpermament voting for its passage. There is one way for the permanent voting members to approve the passage, they all vote for. This leaves at least 4 of the 10 nonpermanent members to vote for its passage to make the requirement of 9 votes. Placing the 1 in front and the 4 or more second we have

1[C(10,4) + C(10,5) + ... + C(10,10)]

which means there 848 ways a measure can be passed.

We will ignore the 1. The easiest way to enter the rest of this long formula into the calculator is hold down the "black up arrow" while you use the "blue left up" to highlight the text shown in the calculator screen shown on the left. Then use the "diamond copy" command; enter "5)"; and then use the "diamond paste" command. Continue with this process until you get all the terms entered.

Problem 13 Drivers' Test - Combination - Addition Principle

"The State Motor Vehicle Department requires learners to pass a written test on the motor vehicle laws of the state. The exam consists of ten true or false questions, of which eight must be answered correctly to qualify for a permit. In how many different ways can a learner who answers all the questions on the exam qualify for a permit?" (Tan 457).

 Online Explanation and Examples 7.4 Permutations and Combinations

Explanation 1 Calculator Screens

You can use the "!" key. It is found above the a's or after the z's in the [CATALOG] or [2nd] Math [7: Probability 1:!]

Working with factorials

Click on the Show 89 movie to see a video of how to enter permutations, factorials, and combinations into our calculator. Example of each are given.

Example 2 Calculator Screen

Working with permutations

Example 3 Calculator Screen

Example 4 Calculator Screen

Example 5 Calculator Screen

Example 6 Calculator Screen

Working with combinations

Example 7 Calculator Screen

Example 8 Calculator Screen

Example 9 Calculator Screen

Example 10 Calculator Screen

Example 11 Calculator Screen

Example 12 Calculator Screen

Example 13 Calculator Screens