Math 221

Integrated Learning SystemWeek 2, Lecture 1

Basic Counting Principles; Permutations and

Combinations

Venn Diagrams

S

A BBA

The symbol AB is read A intersect B, or A and B, and is used to indicate those objects in sets A and B that have the characteristics making them members of both sets.

S

A BBA

S

BAAB’ A’B

A’B’

ExampleS

A BBA

S contains 150 objects. A contains 75 objects; B contains 50 objects and AB contains 20 objects.

1. How many objects are in AB?2. How many objects are in AB’?3. How many objects are in A’B’?4. How many objects are in A’B?

Multiplication Principle

Lets say that you are preparing to cook a meal. You want to serve a meat dish, a starch dish, a vegetable dish and a dessert. You have two different meats in your refrigerator, Three different frozen vegetables, two choices of starches and two different desserts. How many different meals are possible?

The solution to this problem is an example of the multiplication principle.

Let’s say that first you choose a meat. And then you choose a vegetable. Next you choose a starch. An finally, you choose a dessert.

You have two choices at step 1, three choices at step 2, two choices at step 3, and two choices at step 4.

Let’s see how this looks in what we call a decision tree.

Dessert 1Starch 1

Dessert 2Vegetable 1

Dessert 1Starch 2

Dessert 2

Dessert 1Starch 1

Dessert 2Meat 1 Vegetable 2

Dessert 1Starch 2

Dessert 2

Dessert 1Starch 1

Dessert 2Vegetable 3

Dessert 1Starch 2

Dessert 2Start

Dessert 1Starch 1

Dessert 2Vegetable 1

Dessert 1Starch 2

Dessert 2

Dessert 1Starch 1

Dessert 2Meat 2 Vegetable 2

Dessert 1Starch 2

Dessert 2

Dessert 1Starch 1

Dessert 2Vegetable 3

Dessert 1Starch 2

Dessert 2

Example

How many different sequences of five cards can be dealt if you are dealing from a standard deck of 52 cards?

There are 52 ways to deal the first card, 51 ways to deal the second card, 50 ways to deal the third card, 49 ways to deal the fourth card and 48 ways to deal the fifth card.

200,875,3114849505152 Note: This is not the number of possible five card poker hands that can be dealt from a standard deck of 52 cards. In poker hands, the sequence in which the cards are dealt has no effect on the value of the hand.

Factorialsn! (read n factorial) represents the product of the first n natural numbers. It is calculated in this manner:

n! = n(n - 1)(n - 2)...(2)(1)

720123456!6 Examples:

040,9589101112!7

!789101112!7!12

NOTE: 0!=1 by definition

PermutationsA permutation of a set of objects is an arrangement of the objects in a specific order without repetition..Example: How many ways can the five letters a through e be arranged without repetition? (We don’t care whether or not the arrangements spell a word.)There are 5 ways to choose the first letter, 4 ways to choose the second letter, 3 ways to choose the third letter 2 ways to choose the fourth letter and 1 way to choose the fifth letter.

e. - a letters theof nspermutatio 1205!12345 are thereOr,

!

:is , denoted

,repetition without objectsdistinct of nspermutatio ofnumber The

,

,

nP

P

n

nn

nn

Now, suppose we only want to use three of the five letters a - e at a time. How many permutations would we now have?

!35

!5

!2

!5

!2

!2345345

But, ns.permutatio

60345 gives This letter. third thechoose to ways3 andletter

second thechoose to ways4 letter,first thechoose to ways5 are There

!!

:by calculated is and ,by

denoted is timeaat taken objects of nspermutatio ofnumber The

,

,

rn

nP

P

rn

rn

rn

used.often are , and,, symbols the, of placeIn , rnPPPP rnnrrn

Examples

6720

!58!8

5,8

P

200,875,311

!552!52

5,52

P

?9,37 P

?6,21 P

Combinations

A combination of n objects taken r at a time without repetition is a subset of size r of the n objects. The arrangement of the objects does not matter.

We know that the number of permutations of n objects taken r at a time is

!!

, rn

nP rn

But this number takes into account all of the possible arrangements of the r objects. In combinations the different arrangements are not counted as unique. How can we sort of “uncount” the arrangements?

!!

!

! ,

, rnr

n

r

PC rn

rn

. and ,,by denotedoften also is ,

r

nCCC nrrnrn

One way to accomplish this is to divide the number of permutations by the number of arrangements. Remember that the number of arrangements of r objects is r!. Hence the number of combinations of n objects taken r at a time is:

Examples

Now we can answer questions like, what is the number of possible 5 card poker hands that can be dealt from a standard deck of 52 cards? This is the same as, how many combinations of 52 objects taken 5 at a time are there?

860,598,2

!552!5!52

5

52

How many possible 7 card hands that have exactly 3 clubs and 4 spades?

Example

deck. card 52 standard ain suit each of cards 13exactly are There

clubs. 3 draw

to wayspossible 286!313!3

!13

3

13 are thereSo

spades. 4 draw

to wayspossible 715!413!4

!13

4

13 are There

spades. 4 and clubs 3 containing

hands cardseven possible 204,490715286 are thereSo

A catering service offers 8 appetizers, 10 main courses,and 7 desserts. A banquet committee is to select 3 appetizers, 4 main courses, and 2 desserts. How many ways can this be done?

Example for You

An electronics store receives a shipment of 30 graphing calculators, including 6 that are defective. Four of these calculators will be sent to a local high school.

(A) How many selections can be made?

(B) How many of these selections will contain no defective calculators?

Example for You