Counting Principles (Permutations and Combinations )

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  • Counting Principles(Permutations and Combinations )

  • 2012 Pearson Education, Inc.. All rights reserved.

  • Example A combination lock can be set to open to any 4-digitsequence. (a) How many sequences are possible? (b) How many sequences are possible if no digit is repeated?

    2012 Pearson Education, Inc.. All rights reserved. Solution: (a) Since there are 10 digits namely 0, 1, 2..9, thereare 10 choices for each of the digit. By the multiplicationprinciple, there are 10 10 10 10 =10,000 different sequences.

    (b)There are 10 choices for the first digit. It cannot be usedagain, so there are 9 choices for the second digit, 8 choices forthe third digit, and then 7 choices for the fourth digit.Consequently, the number of such sequences is 10 9 8 7 =5040 different sequences.

  • 2012 Pearson Education, Inc.. All rights reserved. ExampleHow many different ways can you choose a bagel, muffin or donut to eat and coffee or juice to drink?Tree diagram

  • Example A teacher is lining up 8 students for a spelling bee. How manydifferent line-ups are possible? 2012 Pearson Education, Inc.. All rights reserved. Solution: Eight choices will be made, one for each space that willhold a student. Any of the students could be chosen for the first space. There are 7 choices for the second space, since 1 studenthas already been placed in the first space; there are 6 choices forthe third space, and so on. By the multiplication principle, the number of different possiblearrangements is 8 7 6 5 4 3 2 1 = 40,320.

  • 2012 Pearson Education, Inc.. All rights reserved. TI-83/84 function (order does matter)

  • Example A teacher wishes to place 5 out of 8 different books on hershelf. How many arrangements of 5 books are possible? 2012 Pearson Education, Inc.. All rights reserved. Solution : The teacher has 8 ways to fill the first space, 7 waysto fill the second space, 6 ways to fill the third, and so onSince the teacher wants to use only 5 books, only 5 spaces canbe filled (5 events) instead of 8, for 8 7 6 5 4 = 6720arrangements.

  • Example Find the number of permutations of the letters L, M, N, O, P,and Q, if just three of the letters are to be used.Solution: 2012 Pearson Education, Inc.. All rights reserved.

  • 2012 Pearson Education, Inc.. All rights reserved. How many permutations are there of two letters from the set {A,B,C}.

  • 2012 Pearson Education, Inc.. All rights reserved. (order does not matter)TI-83/84 function

  • Example How many committees of 4 people can be formedfrom a group of 10 people?Solution: A committee is an unordered group, so use thecombinations formula for C(10,4).

    2012 Pearson Education, Inc.. All rights reserved.

  • 2012 Pearson Education, Inc.. All rights reserved.

  • Example From a class of 15 students, a group of 3 or 4 students will beselected to work on a special project. In how many ways can agroup of 3 or 4 students be selected?Solution: The number of ways to select group of 3 studentsfrom a class of 15 students is C(15, 3) = 455.The number of ways to select group of 4 students from a classof 15 students is C(15, 4) = 1365.The total number of ways to select a group of 3 or 4 studentswill be the 1820. 2012 Pearson Education, Inc.. All rights reserved.

  • 2012 Pearson Education, Inc.. All rights reserved.

  • Example (a) How many 4-digit code numbers are possible if no digits are repeated?Solution: Since changing the order of the 4 digits results in adifferent code, permutations should be used.

    (b) A sample of 3 light bulbs is randomly selected from a batch of 15. How many different samples are possible?

    Solution: The order in which the 3 light bulbs are selected is notimportant. The sample is unchanged if the items are rearranged,so combinations should be used.

    2012 Pearson Education, Inc.. All rights reserved.

  • Example

    (c) In a baseball conference with 8 teams, how many games must be played so that each team plays every other team exactly once?

    Solution: Selection of 2 teams for a game is an unorderedsubset of 2 from the set of 8 teams. Use combinations again. (d) In how many ways can 4 patients be assigned to 6 differenthospital rooms so that each patient has a private room?

    Solution: The room assignments are an ordered selection of 4rooms from the 6 rooms. Exchanging the rooms of any 2patients within a selection of 4 rooms gives a differentassignment, so permutations should be used. 2012 Pearson Education, Inc.. All rights reserved.

  • Introduction to Probability

  • Write the elements belonging to the set {x | x is a state whosename begins with the letter O}.

    2012 Pearson Education, Inc.. All rights reserved. Solution: The state names that begin with the letter O make upthe set {Ohio, Oklahoma, Oregon }.Set a collection of elements that satisfy a certain condition.

  • 2012 Pearson Education, Inc.. All rights reserved. Example 1 Decide if the statement is true or false.

    Solution: The first set is a subset of the second because each element of first set belongs to second set. (The fact that the elements are listed in a different order does not matter.) Therefore, the statement is true.

  • 2012 Pearson Education, Inc.. All rights reserved.

  • 2012 Pearson Education, Inc.. All rights reserved. A) What is the sample space?B) What is the probability of getting a 3?C) What is the probability of getting an odd number?Sample Space The set of all possible outcomes of an experiment or event.

  • ExampleTwo coins are tossed, and a head or a tail is recorded for eachcoin. Write the event E: the coins show exactly one head.

    Solution: Tossing a coin is made up of the outcomes heads (H)or tails (T). If S represents the sample space of tossing twocoins, then S = {HH, HT, TH, TT}.Two outcomes satisfy this condition: so,E = {HT, TH}.

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  • ExampleTwo coins are tossed, and a head or a tail is recorded for eachcoin. Give a sample space for this experiment.What is the probability of getting at least one head.Solution: A) Tossing a coin is made up of the outcomes heads (H)or tails (T). If S represents the sample space of tossing twocoins then S = {HH, HT, TH, TT}. Solution : B) E = {HH,HT,TH} 2012 Pearson Education, Inc.. All rights reserved. Number of elements in the sets

  • http://www.youtube.com/watch?v=mhlc7peGlGg&safe=active Monty Hall probability puzzle

  • Probability of Multiple Events

  • 2012 Pearson Education, Inc.. All rights reserved. 46512(no outcomes in common; if one event occurs the other cannot)(E is called the complement of E)

  • Example

    A study of workers earning the minimum wage grouped such workers into various categories, which can be interpreted as events when a worker is selected at random. Source:Economic Policy Institute. Consider the following events:

    E: worker is under 20;F: worker is white;G: worker is female.

    Describe the following event in words:Solution: 2012 Pearson Education, Inc.. All rights reserved.

  • 2012 Pearson Education, Inc.. All rights reserved. Independent Events The fact that one event has occurred, does not change the probability of the second event occurring.** Events that are mutually exclusive must be dependent, but dependent events are not necessarily mutually exclusive**Read as; the probability of event F occurring given E has occurredRead as; the probability of event E occurring given F has occurred

  • 2012 Pearson Education, Inc.. All rights reserved. (General rule of addition)For mutually exclusive events,So the equation simplifies to

  • Example If a single card is drawn from an ordinary deck of cards, findthe probability of an ace or a club.

    Solution: Let A represent the event an ace and C the eventclub card. There are 4 aces in the deck, so

    There are 13 clubs in the deck, so

    Since there is 1 ace of club in the deck, By the union rule, the probability of the card being an ace or aClub card is 2012 Pearson Education, Inc.. All rights reserved.

  • ExampleSuppose two fair dice are rolled. Find the probability that the sum is 8, or both die show the same number.Solution:

    2012 Pearson Education, Inc.. All rights reserved.

  • ExampleFind the probability that when two fair dice are rolled, the sumis less than 11.

    Solution: To calculate this probability directly, we must find theprobabilities that the sum is 2, 3, 4, 5, 6, 7, 8, 9 or 10 and thenadd them. It is much simpler to first find the probability of the complement, the event that the sum is greater than or equal to 11.

    2012 Pearson Education, Inc.. All rights reserved.

  • (General rule of multiplication)When events are Independent;

    So the equation simplifies to:

  • IndependentNot IndependentIndependentNot Independent1/60.542/79/349/25Not mutually exclusivemutually exclusiveNot mutually exclusive

  • 2012 Pearson Education, Inc.. All rights reserved. P(F); represented by large circle.P(E and F); represented by overlapping section

  • Example

    The following table shows national employment statistics. Use the table to find each probability.8937580143839392972919121

    7.P(male | professional)8.P(laborer | female)9.P(female | sales)10.P(professional | female)11.P(sales | male)12.P(male | laborer)

  • ExampleUse a tree diagram to solve the following problem.

    16.A car insurance company compiled the following information from