Challenge Level

This problem looked at permutations or combinations of possibilities. At first it might seem, as several people thought, the eight runners could each finish the race in just one of three positions -- which they wrote down as 8x3. This gave 24 possible ways of finishing. In fact it is more complex problem than that.

The following solutions show that in fact any one of the eight runners could come first, leaving seven runners who could take the second place and any one of the remaining six could come in third.

The members of **Burgoyne Maths Club** explain
their different approaches to organising the information and how
they began to find a solution by writing down the possible ways of
the runners might finishing the race.

"Some of us gave the 8 pupils names, some letters and some
numbers. We started with listing the combinations with runner
number 1 finishing in first position, and showing where the other
runners could have finished".

123

124

125

126

127

128

132

134

135

136

137

138

142

143

145

146

147

148

152

153

154

156

157

158

162

163

164

165

167

168

172

173

174

175

176

178

182

183

184

185

186

187

Phew!! Writing out solutions really helps you see if all of the
possibilities have been included.

Then the **Burgoyne Maths Club** began to work on
the possibilities if runner number 2 was the one who came in at the
first position: 213

214

215

216

217

218 etc.

"At this point we realised there were 42 positions for each set of results". Once they had discovered a pattern, the Club members were able to finish the problem:

The total possible ways of coming first, second and third must be:

42 x 8 = 336Is this method different? **Bryan Hooi** , of
Primary 5J at Henry Park Primary School described his solution way
of finding the solution.

"There are 8 possibilities for the first place.

For each of these 8, there are the other 7 possibilities for second
place.

Therefore, for first and second place, there are 8 x 7 = 56
possibilities.

For each one of these, there are 6 possibilities for third
place.

Therefore, for first, second and third place, there are 56 x 6 =
__336__ possibilities".

**George Vassilev** from Rosebank Primary School in
Leeds had another method. Can you see how is it similar or
different from the first two ways?

"First I named all the people = 1, 2, 3, 4, 5, 6, 7, 8.Then I
found out how many possible ways can they come in first, second,
and third if the first person is 1. So if the first person is 1
then there are seven people who can be second. So that leaves only
six people to be third. Now I multiplied six, seven, and one
together.

6 * 7 * 1 = 42

That means there are 42 different ways for the people to come in
first, second, and third if the first person is 1. Eight different
people can be first and I have written about one of them but they
all have 42 different ways so that means I times 42 by 8.

42 * 8 = 336

So there are 336 possible ways of the people coming in first,
second, and third places".

**Matthew Tattler** and **Steven
Townend** from Moorfield Junior School explained their
solution differently and in a short way:

"There are eight possible people or ways of coming first. There are seven ways of coming second, and there are six ways of coming third. So that makes the sum 8x7x6, which equals 336, the total number of possible ways".

The same answers were reached by different methods. I wonder if anybody can add an explanation to Matthew and Steven's work that shows how they arrived at their answer.