Robert Yen, Ambarvale High School
In how many ways can 5 boys and 4 girls sit in a row if Katie and Christine want to sit together?
Do these kinds of problems intimidate and confuse you? Permutations and combinations was the one topic that caused me anxiety when I was studying for my HSC. It made little sense to me, and like many students with a probability phobia, I could never be sure if my final answer was correct! And judging by the number of calls to the HSC Advice Line on this very subject, many students today still experience a lack of confidence when using counting techniques to solve probability problems.
To overcome my limitations, I researched and studied this topic thoroughly the first time I taught a 3 Unit class. I knew there had to be a more straightforward, intuitive approach to teaching this topic, hopefully one free of jargon and formula. I'm happy to report that there is, and this became most selfevident some years later on a day when half of my Year 11 class were away on an excursion. I decided to give the remaining half a quick preview of their next 'difficult' topic, but surprised them and myself when I covered most of the permutations and combinations theory in that hour, and without mentioning or .
When I related this story to my head teacher afterwards, he agreed that yes, permutations is best taught using a 'listsandboxes' problemsolving approach, starting from basic principles and relying less upon fancy formulas. What follows are extracts from my teaching notes, which have been refined several times, with the hope that they may help you alleviate some of your students' fears (and perhaps your own) next time you teach this topic.
Introduction
There are five horses in a race  A, B, C, D and E. Suppose you have to bet on which two horses come first and second, in the correct order. From the five horses, how many different bets (1st  2nd pairings) are possible?
AB

BA

CA

DA

EA

AC

BC

CB

DB

EB

AD

BD

CD

DC

EC

AE

BE

CE

DE

ED

Number of arrangements = 20. These arrangements are called permutations.
Now suppose instead that you need to choose the two horses that come first or second, but you don't have to state which one comes first and which once comes second. In other words, the order is not important. Now how many selections are possible?
AB

BC

CD

DE

AC

BD

CE

AD

BE

AE

Number of selections = 10. These selections are called combinations.
Notice how there are only half as many combinations as permutations because order is not important with combinations (e.g., AB is the same as BA).
The multiplication principle for counting arrangements
1.From this list of given names and surnames, how many different arrangements of given name  surname pairs are possible?
Given name

Surname

Alex

Garrett

Brionne

Hinze

Cathy

Ibsen

Darren

Johnson

Erin

King

Fiona


Using a tree diagram, but not completing it ...
For each given name, there are 5 possible surnames.
Total possible arrangements is 6 x 5 = 30.
2. From this menu, calculate how many different 3course dinners are possible.
Entrée

Main course

Dessert

Pumpkin soup

Garlic prawns

Pavlova

Calamari rings

Steak Diane

Black Forest cake

Lasagne

Roast lamb

Chocolate mousse


Chicken Dijon

Mangoes and ice cream


Grilled perch


Total possible dinner arrangements is 3 x 5 x 4 = 60 (and I've tried every one!!).
The basic counting principle is:
1.If A can be arranged in m different ways and B can be arranged in n different ways, then the number of possible arrangements of A and B together is m x n.
2.More generally, if
A can be arranged in a different ways,
B can be arranged in b different ways,
C can be arranged in c different ways, and so on;
then the total number of possible arrangements of ABC ... together is a x b x c ....

Permutations
1. A girls' school is electing a captain and vicecaptain. There are four candidates: Amy, Betty, Caroline and Deane. How many possible arrangements of captain/vicecaptain are there?
Instead of using a tree diagram to count the possibilities, we can draw boxes for the two positions.


There are 4 ways of choosing the captain, but once the captain is determined, there are only 3 ways of choosing the vicecaptain.
Total possible arrangements = 4 x 3 = 12.

These arrangements are called permutations. The number of permutations possible has the notation where n is the number of items or people available, and r is the number of places available. In the above example there were 4 candidates and 2 places.
.
Note is found by multiplying backward from n, r times.
