Problems and Problem Solving - Ministry of Education

Now at first glance, the problem is likely to look daunting – no known method exists and it is difficult
to make guesses when there are as many as 50 clubs, since no basis exists for making these
guesses. However, if we try to simplify the problem by choosing to work with, say, four clubs
(A, B, C, D) we may get further insights which we can apply in solving problems involving larger
numbers. Now, importantly, we have to devise a system for recording how many visits are made
with four clubs visiting each other twice.
Avoid randomly listing possible visits. The list below shows the danger of simply listing combinations
(we are using ‘A B’ to mean ‘A visits B’, while ‘B A’ means that ‘B visits A’).
In the list above, some visits have been repeated while others have been omitted. Instead, try to
be as systematic and organized in listing possible combinations.
In the system outlined below, all the visits that Club A makes are listed in the first row, those that
Club B makes are listed in the second row and so on.
A v B A v C A v D
B v A B v C B v D
C v A C v B C v D
D v A D v B D v C
This system gives us an insight into how the problem can be solved when four clubs exist, each
of them will make one visit to the three other schools giving us a total of 12 visits. An even more
sophisticated system of representation is shown on the following page.
Now that we know that four clubs will make 12 visits, where do we go from here? Does it mean
that the number of visits is three times the number of clubs? Will 99 clubs make 297 visits?
Perhaps before we come to any conclusions, we should try a few more cases. Let us select
carefully what number we try next. Since we know what happens when there are four clubs, let
us explore the number of visits when there are 2, 3 or 5 clubs. This allows us to see any patterns
that might present themselves.
44 PROBLEMS AND PROBLEM SOLVING
ProblemSolving.indd 44
8/24/12 6:55:56 PM