Illinois State University Mathematics Department


MAT 312: Probability and Statistics for Middle School Teachers

Dr. Roger Day (day@math.ilstu.edu)



More Counting Stategies: Combinations


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 we select three items from the 5-element set {A,B,C,D,E} when the order of the three items is disregarded?

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 n items selected r at a time, where the order of selection or the arrangement of the r items is not considered:

and we note that

 




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