Illinois State University Mathematics Department

MAT 312: Probability and Statistics for Middle School Teachers

Spring 1999
9:35 - 10:50 am TR STV 350A
Dr. Roger Day (

Discussion Notes: Session #18
Thursday 18 March 1999

Assignment Due Today
More Counting Stategies: Permulations



Assignment for Next Time

Assignment Due Today
  • Read pages 41 to 51 in Module 3: Counting of the Teacher's Guide textbook.
  • Complete the exercises Starting 3-1 through Starting 3-7 (pages 53 through 59) in the Student Edition textbook.


Here are the three sets of letters we called I, II, and III:

  • Set I: {a,m,r}
  • Set II: {b,d,i,l,u}
  • Set III: {c,e,n,t}

In how many ways can the letters within just one set, from among I, II, and III, be ordered? In set I, we have these possibilities:


We use the Multiplication Principle to describe our selection. We have three letters to choose from in filling the first position, two letters remain to fill the second position, and just one letter left for the last position: 3x2x1=6 different orders are possible. Likewise, for set II there are 120 different ways to order the five letters and there are 24 different ways to order the letters in set III.

Factorial Notation

This above discussion exemplifies the concept of a permutation as an ordered arrangement of items. We also point out the availability of factorial notation to compactly represent the specific multiplication we just carried out: 3x2x1=3!, 5x4x3x2x1=5!, and so on. So n(n-1)(n-2)...(2)(1)=n!.

Example to illustrate use of Permutations:

Almost every morning or evening on the news I hear about the State of Illinois DCFS, the Department of Children and Family Services. I get confused, because our mathematics department has a committee called the Department Faculty Status Committee, or DFSC. Can you see why I'm confused? How many different 4-letter ordered arrangements exist for the set of letters {D, F, S, C}?

Thinking of four positions to fill, __ __ __ __ , we have 4 letters to choose from for the first position, 3 for the next, 2 letters for the next position, and 1 choice for the last position. Using the multiplication principle, there are 4x3x2x1=24 different 4-letter ordered arrangements for the set of letters {D, F, S, C}.

We can extend this application to consider ordered arrangements of only some of the elements in a set. For example, returning to the beverages menu of Blaise's Bistro. If Blaise will post only four possible soda chocies, how many different ordered arrangements of the four sodas are there?

Thinking of four positions to fill, __ __ __ __, we have 6 sodas to choose from for the first position, 5 for the next, 4 sodas for the next, and 3 sodas for the last position. Using the multiplication principle, there are 6x5x4x3=360 different ways to select and order four of the six sodas on the menu.

In general, we use the notation P(n,r) to represent the number of ways to arrange r objects from a set of n objects. In the first problem above, we determined that P(4,4)=24, and in the second we calculated P(6,4)=360. The general value of P(n,r) is n(n-1)(n-2)...([n-(r-1)] or P(n,r)=n(n-1)(n-2)...(n-r+1). Note that n can be any nonnegative integer. Are there any restrictions on the value of r?

There is a step of arithmetic we can apply to the general pattern for P(n,r) to help streamline permutation calculations. In the second line below, we have multiplied by , which is just the value 1 because the numerator and denominator are equal. In the fourth line below we see how the expression can be simplified using factorial notation.

Thus, we have P(6,2)=6!/4! And P(40,8)=40!/32!.

What about P(4,4)? The result above suggests P(4,4)=4!/0!. We already know that P(4,4)=4x3x2x1=4!, So we have 4!=4!/0!. For this to be true, it must be the case that 0!=1. As strange as that may appear, we need 0!=1 in order to maintain consistency within the calculations we wish to carry out.

Assignment for Next Time