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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