12.5 CHAPTER 12. COMBINATIONS AND PERMUTATIONS
will have a total of n 1 × n 2 possible outcomes. Thisidea can be generalisedto m experiments as the
total number of outcomes for m experiments is:
n 1 ×n 2 ×n 3 ×···×nm=
�m
i=1
ni
�
is the multiplication equivalent of
�
Note: the order in which the experiments are done does not affect thetotal number of possibleout-
comes.
Example 1: Lunch Special
QUESTION
A take-away has a 4-piece lunch special which consists of a sandwich, soup, dessert and drink
for R 25 , 00. They offer the following choices for :
Sandwich: chicken mayonnaise, cheese and tomato, tuna, and ham andlettuce
Soup: tomato, chicken noodle, vegetable
Dessert: ice-cream, piece of cake
Drink: tea, coffee, coke,Fanta and Sprite.
How many possible meals are there?
SOLUTION
Step 1 : Determine how many parts to the meal there are
There are 4 parts: sandwich, soup, dessert and drink.
Step 2 : Identify how many choices there are for each part
Meal component Sandwich Soup Dessert Drink
Number of choices 4 3 2 5
Step 3 : Use the fundamental counting principle to determine how many different meals
are possible
4 × 3 × 2 × 5 = 120
So there are 120 possible meals.
12.5 Combinations
EMCDI
The fundamental counting principle describeshow to calculate the total number of outcomeswhen
multiple independent events are performed together.