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  • MDM4U-B

    Combinations

    7

  • Mathematics of Data Management MDM4U-B Lesson 7 1

    Introduction In Lesson 6, you learned about permutations and how to solve counting problems where the order of selection matters. In this lesson, you will learn to solve counting problems where order does not matter. There are many situations in which youre not interested in the order of choosing the itemsyoure only interested in selecting a certain number of them. Here are some examples:

    Thenumberofcombinationsoftoppingsyoucanhaveonyourfavouritefast-foodmeal

    Thenumberofcombinationsinsomelotterygameswhereyouchoosesixorsevennumbers out of 49

    Thenumberofcombinationsofoptionsincellphoneplans

    Thenumberofsix-personweddingpartiesthatcanbechosenfrom20friends

    For fun, you can use the formulas in this lesson to calculate your chances of winning first prize in the lottery!

    Planning Your Study You may find this time grid helpful in planning when and how you will work through this lesson.

    Suggested Timing for This Lesson (Hours)Combinations Counting Combinations 1Combination Problems 1Solving Equations with Combinations Key Questions 1

    What You Will LearnAfter completing this lesson, you will be able to

    recognizesituationsinvolvingcombinations

    solvecountingproblemsinvolvingcombinations

    solvecombinationequations

    Copyright 2009 The Ontario Educational Communications Authority. All rights reserved.www.ilc.org

  • Lesson 7 Mathematics of Data Management MDM4U-B2

    Combinations In the previous lesson, you looked at permutations, which were defined as arrangements of objects where order is important (for example, where A,B,C was considered to be different from A,C,B). In the current lesson, you will concentrate on combinations of objects where order is not important.

    Just as with the counting problems in Lesson 6, if you have a problem involving a small number of choices, then its easy to answer the problem by making an exhaustive list of all the possibilities. However, once the numbers start to get a little bigger, it becomes very time-consumingtomakeacompletelistanditseasiertouseformulas.

    Here is an example of a combination problem using small numbers that shows you how the general formula is derived from the complete list of outcomes.

    Example:

    Suppose that you want to know how many ordered pairs can be formed by selecting two of the letters A,B,C,D, where you can use a letter only once. In other words, you want to count the number of permutations of two of those four letters.

    Solution:

    In the previous lesson, you already saw that the number of such permutations is:

    P(4,2) = 4!

    (4 2)!=

    4 3 2121

    = 4 3 = 12

    The numbers in this example are small enough for you to list all of the possible permutations in a table:

    AB BA CA DAAC BC CB DBAD BD CD DC

    Now, suppose youre not interested in counting permutations, but want to count combinations instead. In other words, youre now saying that order doesnt matter, so A,B is the same as B,A; A,C is the same as C,A; and so on.

    This example is small enough to allow you to list all of the possible combinations in this case:

    A,B (which is the same as B,A)

    A,C (which is the same as C,A)

    A,D (which is the same as D,A)

    B,C (which is the same as C,B)

    B,D (which is the same as D,B)

    C,D (which is the same as D,C)

    Copyright 2009 The Ontario Educational Communications Authority. All rights reserved. www.ilc.org

  • Mathematics of Data Management MDM4U-B Lesson 7 3

    In this example, the number of combinations is equal to 6. This is equal to the number ofpermutationsdividedby2.Anotherwaytolookatthisexampleisthatthenumberofcombinationsisequaltothenumberofpermutationsdividedby2!.

    What happens if you compare the number of permutations and combinations of three letters chosen from A,B,C,D? You know the total number of permutations of three letters from A,B,C,D is P(4,3)=432=24.Thatsstillsmallenoughforyoutobeabletolistall of the permutations in a table:

    ABC BAC CAB DABABD BAD CAD DACACB BCA CBA DBAACD BCD CBD DBCADB BDA CDA DCAADC BDC CDB DCB

    The number of possible combinations will be less than the number of possible permutations, since several permutations will correspond to the same combination.

    It is possible to list all of the possible combinations of three of the letters A,B,C,D:

    A,B,C (which is the same as A,C,B or B,A,C or B,C,A or C,A,B or C,B,A)

    A,B,D (which is the same as A,D,B or B,A,D or B,D,A or D,A,B or D,B,A)

    A,C,D (which is the same as A,D,C or C,A,D or C,D,A or D,A,C or D,C,A)

    B,C,D (which is the same as B,D,C or C,B,D or C,D,B or D,B,C or D,C,B)

    In this example, the number of combinations is equal to 4. This is equal to the number of permutations divided by 6. Another way to look at this example is that the number of combinations is equal to the number of permutations divided by 3! since 6 is equal to 3!. The reason for using the factorial notation will become clear in the explanation that follows.

    In the first example given above, you should notice:

    (Number of combinations of two letters from A,B,C,D) = (number of permutations of two lettersfromA,B,C,D)2!

    and in the second example:

    (Number of combinations of three letters from A,B,C,D) = (number of permutations of three letters from A,B,C,D) 3!

    Do you recognize a pattern illustrated by the above examples? The general rule is that the number of combinations is equal to the number of permutations divided by the factorial of the number of choices.

    Forexample,inthecaseofhavingthreechoices,thereareatotalof432=24different permutations of three letters from A,B,C,D. But if its combinations you care about, then you gather the various permutations into groups, where each group consists of 3! = 6 permutations. The permutations in a group all correspond to the same

    Copyright 2009 The Ontario Educational Communications Authority. All rights reserved.www.ilc.org

  • Lesson 7 Mathematics of Data Management MDM4U-B4

    combination. The reason why each group has 3! = 6 permutations in this example is that a set of three elements can be arranged in 3! = 6 different ways.

    In Lesson 6, you were introduced to the notation P(n,r), which stands for the number of possible permutations of r elements chosen from a set of n elements. Its also useful to have a notation for the number of possible combinations of r elements chosen from a set of n elements. That number is denoted as C(n,r).

    Counting Combinations Combinations are a selection of items where order does not matter. The number of combinations of r items from a set of n objects is given by the formula:

    C(n,r) = P(n,r)

    r !Recall from the previous lesson that you learned that

    P(n,r) = n!

    (n r)!

    This means that our above formula for C(n,r) can also be written as:

    C(n,r) = n!

    (n r)!r!This is called the factorial form.

    There are other ways to write the symbol for a combination problem. Instead of C(n,r),

    sometimes you might see it written as nCr or

    nr

    . You will mainly use the common

    notation C(n,r) throughout this unit, but its helpful to be able to recognize the other forms. You pronounce all of these expressions as n choose r, since it stands for the number of ways of choosing an (unordered) set of r elements from a set of n elements.

    When it comes to computing C(n,r) for specific values of n and r, its usually simpler to

    use the formula P(n,r)

    r ! since the numerator and denominator are both products of r

    positive integers. Here are two examples using this formula:

    C(7,3) = P(7,3)

    3!=

    7 653 21

    = 35

    C(23,5) = P(23,5)

    5!=

    23 22 21 20 195 4 3 2 1 = 33 649

    For small numbers of n and r, its easy to work out the factorials and come up with a numerical answer. However, when the numbers start to get bigger, it can be awkward to calculate large factorials without a computer. For this reason, its often easier to leave the answer in factorial form, as in the following example:

    C(42,13) = 42!

    (42 13)!13!

    Copyright 2009 The Ontario Educational Communications Authority. All rights reserved. www.ilc.org

  • Mathematics of Data Management MDM4U-B Lesson 7 5

    Example:

    Thereare10membersintheMiltondebateclub.Thecoachwantstoselectthreemembers from the debate club to form a team to attend the national championships in Ottawa. How many ways can the coach make her choice?

    Solution:

    Youwanttoselectthreemembersfrom10whereorderisunimportant,soyouaretalkingabout combinations here. The number of possible combinations of three elements chosen fromasetof10is:

    C(10,3) = P(10,3)

    3!

    =

    10 9 83 21

    =720

    6= 120

    So,thecoachcanselecttheteamin120ways.

    Example:

    From a selection of six playing cards, how many ways can you choose

    a) three cards?

    b) six cards?

    Solution:

    a) If you want to choose three cards from the six cards, then you are making an unordered choice of three items from six items. In other words, you want to count the number of possible combinations of three of the six items.

    The number of such combinations is:

    C(6,3) = P(6,3)

    3!=

    65 43 21

    = 20

    So,thereare20waystoselectthreecardsoutofthesix.

    b) The number of possible combinations of six of the six cards is:

    C(6,6) = 6!

    6!(6 6)!=

    6!6! 0!

    = 1

    This makes sense, since there is only one possible way to select six cards out of six.

    Example:

    Show that:

    103

    +

    104

    =

    114

    C