CHAPTER 4: COUNTING

TECHNIQUES

Mr. Mark Anthony Garcia, M.S.

Mathematics Department

De La Salle University

Experiment: Definition

An experiment is a process that generates a set of data.

Example 1: Experiment

Tossing of a coin Tossing of a coin twice Rolling a pair of dice Drawing a card from a deck of playing

cards

Sample Space

The set of all possible outcomes of a statistical experiment is called the sample space and is denoted by S. An element of a sample space is called a sample point.

Example 2: Sample Space

Consider the experiment of flipping a coin. If a head occurs, we flip the coin for the second time while if a tail occurs, a die is thrown.

S = {HH, HT, T1, T2, T3, T4, T5, T6 }

Example 3: Sample SpaceSuppose that three balls are selected one at a time from a bag containing white and black balls. Let B represents a black ball and W a white ball. The sample point BBW is obtained from selecting a black ball on the first and second selection and a white ball on the third selection. S = {BBB, BBW, BWB, BWW, WBB,

WBW, WWB, WWW}

Event

An event is a subset of a sample space.

Example 4: Event

Consider the experiment of tossing a coin.

Then S = {H, T}. Let E be the event of obtaining a head. Then E = {H}.

Example 5: Event

Consider the experiment of tossing a coin twice.

Then S = {HH, HT, TH, TT}. Let E be the event of obtaining exactly

two heads. Then E = {HH}.

Example 6: Event

Consider the experiment of rolling a pair of dice.

Let E be the event of obtaining a “double”.

Then E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}.

Example 7: Event

Consider the experiment of drawing a card from a deck.

Let E be the event of obtaining a heart card.

Then E = {AH, 2H,…,10H, JH, QH, KH}

Counting Techniques

In probability, one needs to know how to count the number of elements of a sample space and the number of elements of an event.

To do this, we use the concept of the fundamental principle of counting, permutations and combinations.

Fundamental Principle of Counting: Multiplication Rule

If an event M can occur in m ways and event N can occur in n ways, then event M followed by event N can occur in mn ways.

Example 8: Multiplication Rule

A certain shoe comes in 5 different styles with each style available in 4 distinct colors. If the store wishes to display pairs of these shoes showing all of its various styles and colors, how many different pairs would the store have on display?

Example 9: Multiplication Rule

In a hospital, patients are grouped according to their gender (M, F) and blood types (A, B, AB, O). How many groups of patients are possible?

Generalized Multiplication Rule

If an event can occur in ways, event can occur in ways,…, event can occur in ways, then event followed by event up to event can occur in ways.

Example 10: Generalized Multiplication Rule

A developer of a new subdivision offers a prospective home buyer a choice of 4 designs, 3 different heating systems, a garage or carport, and a patio or screened porch. How many different plans are available to the buyer?

Example 11: Generalized Multiplication Rule

The digits 1, 2, 3, 4 and 5 are to be used in a 4-digit ID card.  How many different cards are possible if repetitions are permitted?

Example 12: Generalized Multiplication Rule

The digits 1, 2, 3, 4 and 5 are to be used in a 4-digit ID card.  How many different cards are possible if repetitions are not permitted?

(5) (4) (3) (2) 120

Example 13: Generalized Multiplication Rule

How many three digit numbers can be formed from the digits 0 to 9 that are greater than 600 if digits cannot be used more than once?

(4) (9) (8) 288

Permutations

Permutation is the process of arranging objects in a linear or circular manner. In permutations, ordering is important. For example, the ordering (green, blue, red) is different from the ordering (red, blue, green).

Situation: Permutations

Consider three objects A, B and C and arrange them in a linear manner.

Exhaust all the possible arrangements of all the three objects in a linear manner.

Situation: Permutations

A B C

B A C

C A B

A C B

B C A

BC A

Situation: Permutations

There are six possible permutations of the three objects A, B and C. Using the generalized multiplication rule, we have

(3) (2) (1) 6

Permutation Formula: Arranging all n distinct objects

The number of ways of arranging n distinct objects all at a time is given by the formula

Note that and .

Example 14: PermutationHow many different permutations or arrangements can be formed from the following words if the letters are taken all at a time? A. FORMAT, B. PICTURE

A. 6! B. 7!

Example 15: PermutationIn how many different ways can the curator of a museum arrange five paintings next to each other horizontally on a wall?

5! = 120

Example 16: Permutation within a permutation

The boys and their respective girlfriends have bought six tickets for a concert. In how many different ways can these people be seated in a row if

A. there are no restrictions on the sitting arrangement?

B. each couple is to sit together?

C. all the boys are to sit together and all the girls are to sit together?

Example 17: Permutation within a permutation

In how many ways can four different Math books, two different English books, and three Psychology books be arranged on a shelf so that all books in the same subjects are together?

Using generalized multiplication rule, we have 4!2!3!3! = 1728

Situation: Permutations

Consider four objects A, B, C and D. Take three objects at a time. Exhaust all the possible arrangements

of the four objects taking three objects at a time.

Situation: Permutations

ABC ABD ACD BCDACB ADB ADC BDCBAC BAD CAD CBDBCA BDA CDA CDBCAB DAB DAC DBCCBA DBA DCA DCB

Situation: Permutations

There are 24 three-object permutations taken from four objects.

Using generalized multiplication rule, we have (4)(3)(2) = 24 or we use the permutation formula 4P3 = 24.

Permutation Formula: n distinct objects taken r objects at a time

The number of ways of arranging n distinct objects taking r objects at a time is given by

Example 18: Permutations

In how many ways can 5 starting positions on a basketball team be filled with 8 men who can play any of the positions?

Example 19: Permutations

If five sprinters are competing in a race during a track meet, how many arrangements are possible for first, second, and third place finishers?

Using generalized multiplication rule, we have (5)(4)(3) = 60 arrangements.

Using permutation, we have 5P3 = 60.

Combinations

Combination is the process of selecting r objects from a group of n objects. In combinations, ordering is not important. For example, the ordering (green, blue, red) is same to the ordering (red, green, blue).

Situation: Combinations

Consider four objects A, B, C and D. Take three objects at a time. Exhaust all three-object combinations

from the four objects.

ABC ABD ACD BCD

Situation: Combinations

There are 4 three-object combinations from four objects.

This is obtained from the combination formula 4C3.

Combination Formula

The number of ways of selecting r objects from n distinct objects is given by the formula

Example 20: CombinationsIn how many ways can three class representatives be chosen from a group of twelve students? 

12C3 = 220

Example 21: CombinationsIn the Philippine 6/42 lotto system, wherein a bet consists of choosing any 6 numbers (in no particular order) from 1 to 42, how many possible bets can be made?

5,245,786

Example 22: CombinationsA bag contains four white balls, six yellow balls, and five red balls. In how many ways can six balls be chosen if there must be two balls of each color?

4C2 = 6

6C2 = 15

5C2 = 10

Example 22: Combinations

Combinations for white: 4C2 = 6 Combinations for yellow: 6C2 = 15 Combinations for red: 5C2 = 10 Using generalized multiplication rule, we

have (6)(15)(10) = 900 combinations