### Calendar Capers

Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat this for a number of your choice from the second row. You should now have just one number left on the bottom row, circle it. Find the total for the three numbers circled. Compare this total with the number in the centre of the square. What do you find? Can you explain why this happens?

### Reverse to Order

Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number?

### Card Trick 2

Can you explain how this card trick works?

# Cross-country Race

##### Stage: 3 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 = 336

Is 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.