Illinois State University Mathematics Department

MAT 312: Probability and Statistics for Middle School Teachers Dr. Roger Day (day@ilstu.edu) |

More Counting Stategies: Combinations |
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Combinations

What is the distinction between asking these two questions?(i) In how many ways can a 5-card poker hand be dealt?(ii) How many different 5-card poker hands exist?

The first question considers the order or arrangement of the cards as they are dealt. In the second question, the end result when dealt 2H,4D,JC,3S,10D in that order is the same as being dealt 4D,3S,JC,10D,2H in that order. In each case, the same 5-card poker hand exists.

Using permutations, we found P(52,5) as the solution to the first problem. That is, we arranged 5 objects selected from among 52 cards. For the second question, called a

combination, there are many arrangements that yield the same 5-card hand. We need to account for this. Let's consider a simpler problem.How many ordered arrangements exist for the letters of the set {A,B,C,D,E}?

Using permutations, we have P(5,5) = 5! = 120 ways to arrange the five letters.

How many ordered arrangements are there of 3 items from the 5-element set?

We have P(5,3) = 543 = 5!/2! = 60 arrangements. For example, for the three letters {A,B,C} we have these arrangements: ABC, ACB, BAC, BCA, CAB, CBA. This represents 6 of the 60 arrangements, yet each involves the same selection of three letters. Likewise for the three letters {A,C,E}: We have ACE, AEC, CAE, CEA, EAC, ECA.

It seems that for each 3-letter subset of {A,B,C,D,E} there are 6 arrangements of the same three letters. This is a helpful observation in exploring the following question:

How many ways can weselectthree items from the 5-element set {A,B,C,D,E} when theorderof the three items isdisregarded?One way is to list the unique 3-element subsets of {A,B,C,D,E}: ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE. There are 10 such 3-element subsets.

Here's another way to consider the count:

(i) There are P(5,3) = 60 ordered arrangements of the 5-element set into 3-element subsets.(ii) Within the 60 ordered arrangements, there are 10 groups, each with 6 arrangements that use the same 3-letter subset. That is, 60 ÷ 6 = 10 unique 3-element subsets. Using combinatorics notation, we have

In general, we have a way to determine the number of combinations of

nitems selectedrat a time, where the order of selection or the arrangement of theritems is not considered:

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