# PERMUTATIONS AND COMBINATIONS BOTH PERMUTATIONS AND COMBINATIONS USE A COUNTING METHOD CALLED FACTORIAL

• View
222

• Download
0

Embed Size (px)

### Text of PERMUTATIONS AND COMBINATIONS BOTH PERMUTATIONS AND COMBINATIONS USE A COUNTING METHOD CALLED...

• PERMUTATIONS AND COMBINATIONS

• BOTH PERMUTATIONS AND COMBINATIONSUSE A COUNTING METHOD CALLED FACTORIAL.

• A FACTORIAL is a counting method that uses consecutive whole numbers as factors.The factorial symbol is !Examples 5! = 5x4x3x2x1 = 120 7! = 7x6x5x4x3x2x1 = 5040

• First, well do some permutation problems.

Permutations are arrangements.

• PermutationsIn the context of counting problems, arrangements are often called permutations; the number of permutations of n things taken r at a time is denoted nPr. Applying the fundamental counting principle to arrangements of this type givesnPr = n(n 1)(n 2)[n (r 1)].

• Factorial Formula for PermutationsThe number of permutations, or arrangements, of n distinct things taken r at a time, where r n, can be calculated as

• Example: PermutationsEvaluate each permutation.a) 5P3b) 6P6Solution

• Example: IDsHow many ways can you select two letters followed by three digits for an ID if repeats are not allowed? SolutionThere are two parts:1. Determine the set of two letters.2. Determine the set of three digits.Part 1 Part 2

• Example: Building Numbers From a Set of DigitsHow many four-digit numbers can be written using the numbers from the set {1, 3, 5, 7, 9} if repetitions are not allowed?SolutionRepetitions are not allowed and order is important, so we use permutations:

• Lets do a permutation problem.How many different arrangements are there for 3 books on a shelf?Books A,B, and C can be arranged in these ways:ABC ACB BAC BCA CAB CBASix arrangements or 3! = 3x2x1 = 6

• In a permutation, the order of the books is important.Each different permutation is a different arrangement.The arrangement ABC is different from the arrangement CBA, even though they are the same 3 books.

• You try this one:1. How many ways can 4 books be arranged on a shelf?4! or 4x3x2x1 or 24 arrangementsHere are the 24 different arrangements:ABCD ABDC ACBD ACDB ADBC ADCBBACD BADC BCAD BCDA BDAC BDCACABD CADB CBAD CBDA CDAB CDBADABC DACB DBAC DBCA DCAB DCBA

• Now were going to do 3 books on a shelf again, but this time were going to choose them from a group of 8 books.Were going to have a lot more possibilities this time, because there are many groups of 3 books to be chosen from the total 8, and there are several different arrangements for each group of 3.

• If we were looking for different arrangements for all 8 books, then we would do 8!But we only want the different arrangements for groups of 3 out of 8, so well do a partial factorial, 8x7x6=336

• Try these:1. Five books are chosen from a group of ten, and put on a bookshelf. How many possible arrangements are there?10x9x8x7x6 or 30240

• 2. Choose 4 books from a group of 7 and arrange them on a shelf. How many different arrangements are there?7x6x5x4 or 840

• Combinations

• CombinationsIn the context of counting problems, subsets, where order of elements makes no difference, are often called combinations; the number of combinations of n things taken r at a time is denoted nCr.

• Factorial Formula for CombinationsThe number of combinations, or subsets, of n distinct things taken r at a time, where r n, can be calculated as Note:

• Example: CombinationsEvaluate each combination.a) 5C3b) 6C6Solution

• Now, well do some combination problems.

Combinations are selections.

• There are some problems where the order of the items is NOT important.These are called combinations.You are just making selections, not making different arrangements.

• Example: A committee of 3 students must be selected from a group of 5 people. How many possible different committees could be formed?Lets call the 5 people A,B,C,D,and E.Suppose the selected committee consists of students E, C, and A. If you re-arrange the names to C, A, and E, its still the same group of people. This is why the order is not important.

• Because were not going to use all the possible combinations of ECA, like EAC, CAE, CEA, ACE, and AEC, there will be a lot fewer committees.Therefore instead of using only 5x4x3, to get the fewer committees, we must divide.5x4x33x2x1(Always divide by the factorial of the number of digits on top of the fraction.)Answer: 10 committees

• Now, you try.1. How many possible committees of 2 people can be selected from a group of 8? 8x72x1or 28 possible committees

• 2. How many committees of 4 students could be formed from a group of 12 people?12x11x10x9 4x3x2x1or 495 possible committees

• Example: Finding the Number of SubsetsFind the number of different subsets of size 3 in the set {m, a, t, h, r, o, c, k, s}.SolutionA subset of size 3 must have 3 distinct elements, so repetitions are not allowed. Order is not important.

• Example: Finding the Number of Poker HandsA common form of poker involves hands (sets) of five cards each, dealt from a deck consisting of 52 different cards. How many different 5-card hands are possible?SolutionRepetitions are not allowed and order is not important.

• Guidelines on Which Method to Use

PermutationsCombinationsNumber of ways of selecting r items out of n itemsRepetitions are not allowedOrder is important.Order is not important.Arrangements of n items taken r at a timeSubsets of n items taken r at a timenPr = n!/(n r)!nCr = n!/[ r!(n r)!]Clue words: arrangement, schedule, orderClue words: group, sample, selection

• Permutations and CombinationsEvaluate each problem.c) 6P6a) 5P3b) 5C3d) 6C654360107201

• Permutations and CombinationsHow many ways can you select two letters followed by three digits for an ID if repeats are not allowed? Two parts:2. Determine the set of three digits.1. Determine the set of two letters.26P210P326256501098720650720468,000

• Permutations and CombinationsA common form of poker involves hands (sets) of five cards each, dealt from a deck consisting of 52 different cards. How many different 5-card hands are possible?Hint: Repetitions are not allowed and order is not important.52C52,598,9605-card hands

• Permutations and CombinationsFind the number of different subsets of size 3 in the set: {m, a, t, h, r, o, c, k, s}.Find the number of arrangements of size 3 in the set: {m, a, t, h, r, o, c, k, s}.9C384Different subsets9P3987504arrangements

• 10.3 Using Permutations and CombinationsGuidelines on Which Method to Use

• CIRCULAR PERMUTATIONSWhen items are in a circular format, to find the number of different arrangements, divide: n! / n

• Six students are sitting around a circular table in the cafeteria. How many different seating arrangements are there?6! 6 = 120

• Fundamental Counting PrincipleFor a college interview, Robert has to choose what to wear from the following: 4 slacks, 3 shirts, 2 shoes and 5 ties. How many possible outfits does he have to choose from? 4*3*2*5 = 120 outfits

• A combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated?

• PermutationsA combination lock will open when the right choice of three numbers (from 1 to 30, inclusive) is selected. How many different lock combinations are possible assuming no number is repeated? Practice:

• From a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled?

• PermutationsFrom a club of 24 members, a President, Vice President, Secretary, Treasurer and Historian are to be elected. In how many ways can the offices be filled? Answer:

• To play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible?

• CombinationsTo play a particular card game, each player is dealt five cards from a standard deck of 52 cards. How many different hands are possible? Answer:

• A student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions?

• CombinationsA student must answer 3 out of 5 essay questions on a test. In how many different ways can the student select the questions? Answer:

• A basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards?

• CombinationsA basketball team consists of two centers, five forwards, and four guards. In how many ways can the coach select a starting line up of one center, two forwards, and two guards? Answer:Thus, the number of ways to select the starting line up is 2*10*6 = 120.

• How many ways can a student government select a president, vice president, secretary, and treasurer from a group of 6 people?This is the equivalent of selecting and arranging 4 items from 6.= 6 5 4 3 = 360Divide out common factors.There are 360 ways to select the 4 people.

• How many ways can a stylist arrange 5 of 8 vases from left to right in a store display?Divide out common factors. = 8 7 6 5 4= 6720There are 6720 ways that the vases can be arranged.

• Awards are given out at a costume party. How many ways can most creative, silliest, and best costume be

Recommended ##### L14: Permutations, Combinations and Some L14: Permutations, Combinations and Some Review EECS 203: Discrete
Documents ##### Section 13.2 – Permutations and Combinations .281 Section 13.2 – Permutations and Combinations
Documents ##### Permutations and Combinations - ACU Blogs Permutations and Combinations Section 13.3-13.4 Dr. John Ehrke
Documents ##### Permutations and Combinations AII.12 2009. Objectives: apply fundamental counting principle compute permutations compute combinations distinguish permutations
Documents ##### Permutations and Combinations Standards: MM1D1b. Calculate and use simple permutations and combinations
Documents ##### BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS CHAPTER
Documents Documents