section 11.1

Section 11.1The Fundamental Counting Principle and Permutations

Math in Our World

Learning Objectives

Use the fundamental counting principle. Calculate the value of factorial expressions. Find the number of permutations of n objects. Find the number of permutations of n objects

taken r at a time. Find the number of permutations when some

objects are alike.

Fundamental Counting PrincipleIn a sequence of n events in which the first event can occur in k1 ways and the second event can occur in k2 ways and the third event can occur in k3 ways and so on, the total number of ways the sequence can occur is

k1 k2 k3 . . . kn

EXAMPLE 1 Using the Fundamental Counting Principle

There are four blood types: A, B, AB, and O. Blood is also either Rh+ or Rh-. If a local blood bank labels donations according to type, Rh factor, and gender of the donor, how many different ways can a blood sample be labeled?

EXAMPLE 1 Using the Fundamental Counting Principle

There are four blood types: A, B, AB, and O. Blood is also either Rh+ or Rh-. If a local blood bank labels donations according to type, Rh factor, and gender of the donor, how many different ways can a blood sample be labeled?

There are four possibilities for blood type, two for Rh factor, and two for gender of the donor. Using the fundamental counting principle, there are

4 x 2 x 2 = 16different ways that blood could be labeled.

SOLUTION

Ex: Permutation: Arrange books

EXAMPLE 2 Using the Fundamental Counting Principle w/Repetition

(a) The letters A, B, C, D, and E are to be used in a four-letter ID card. How many different cards are possible if letters are allowed to be repeated?

(b) How many cards are possible if each letter can only be used once?

EXAMPLE 2 Using the Fundamental Counting Principle w/Repetition

SOLUTION(a) There are four spaces to fill and five choices for each. The fundamental counting principle gives us

5 x 5 x 5 x 5 = 54 = 625(b) The first letter can still be chosen in five ways. But with no repetition allowed, there are only four choices for the second letter, three for the third, and two for the last. The number of potential cards is

5 x 4 x 3 x 2 = 120Hopefully, this ID card is for a pretty small organization.

Factorial NotationFor any natural number n

n! = n(n – 1)(n – 2)(n – 3) . . . 3 x 2 x 1n! is read as “n factorial.”0! is defined as 1.

The symbol for a factorial is the exclamation mark (!). In general, n! means to multiply the whole numbers from n down to 1. For example,

1! = 1 = 1 3! = 3 x 2 x 1 = 62! = 2 x 1 = 2 4! = 4 x 3 x 2 x 1 = 24

Factorial NotationSome of the formulas we’ll be working with require division of factorials. This will be simple if we make two key observations:

• You can write factorials without writing all of the factors down to 1. For example,

5! = 5 x 4 x 3 x 2 x 1, but we can also write this as 5 x 4!, or 5 x 4 x 3!, etc.

EXAMPLE 3 Evaluating Factorial Expressions

Evaluate each expression:(a) 8! (b)12!

10!

EXAMPLE 3 Evaluating Factorial Expressions

Evaluate each expression:(a) 8! (b)

SOLUTION(a) 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320(b) First, write 12! as 12 x 11 x 10!, then note that

10!1

10!

12! 12 11 10! 12 11 13210! 10!

12!10!

Permutations

An arrangement of n distinct objects in a specific order is called a permutation of the objects.

The number of permutations of n distinct objects using all of the objects is n!.

EXAMPLE 4 Calculating the Number of Permutations

In seven of the 10 years from 1998–2007, the five major league baseball teams in the American League East division finished in the exact same order: New York, Boston, Toronto, Baltimore, Tampa Bay. Just how unusual is this? Find the number of possible finishing orders for these five teams.

EXAMPLE 4 Calculating the Number of Permutations

SOLUTIONThis is a permutation problem since we’re deciding on the number of ways to arrange five distinct objects. There are

5! = 120

possible finishing orders.

EXAMPLE 5 Solving a Permutation Problem

How many different ways can a pledge class with 20 members choose a president, vice president, and Greek Council representative? (No pledge can hold two offices.)

EXAMPLE 5 Solving a Permutation Problem

How many different ways can a pledge class with 20 members choose a president, vice president, and Greek Council representative? (No pledge can hold two offices.)SOLUTIONThere are 20 choices for president, 19 remaining candidates for vice president, and 18 members left to choose from for Greek Council rep.So there are 20 x 19 x 18 = 6,840 different ways to assign these three offices.

Permutations

Permutation of n Objects Taken r at a TimeThe arrangement of n objects in a specific order using r of those objects is called a permutation of n objects taken r at a time. It is written as nPr , and is calculated using the formula

!( )!n r

nPn r

EXAMPLE 6 Solving a Permutation Problem

How many five-digit zip codes are there with no repeated digits?

SOLUTIONThis is a permutation problem because five numbers are taken from 10 possible digits, with order important and no repetition. In this case, n = 10 and r = 5.

10 510!

(10 5)!P

10!5!

10 9 8 7 6 5!

5!

30,240

Permutations

Permutation Rule When Objects Are AlikeThe number of permutations of n objects in which k1 objects are alike, k2 objects are alike, etc. is

1 2

!! !... !p

nk k k

where k1 + k2 + . . . + kp = n

EXAMPLE 7 Solving a Permutation Problem with Like Objects

How many different passwords can be made using all of the letters in the word Mississippi ?

SOLUTIONThe letters can be rearranged as M IIII SSSS PP. Then n = 11, k1 = 1, k2 = 4, k3 = 4, and k4 = 2.Using our newest formula, there are

11!1!4!4!2!

34,650

different passwords.