Imagine a three dimensional version of noughts and crosses
where two players take it in turn to place
different coloured marbles into a box.
The box is made from 27 transparent unit cubes arranged in a 3-by-3-by-3 array.
The object of the game is to complete as many winning lines of three marbles as possible.
How many different winning lines are there?
Once you have thought about how you might tackle the problem, click below to see four different methods for working out the number of winning lines in a $3 \times 3 \times 3$ cube.
From a vertex there are $7$ other vertices that you can join to in order to make a winning line. $7 \times 8 = 56$ lines, but this counts each line from both ends, so there are $28$ 'vertex' winning lines.
From the middle of an edge there are $3$ other middles-of-edges that you can join to in order to make a winning line. $3 \times 12 = 36$ lines, but this counts each line from both ends so there are $18$ 'middle of edge' winning lines.
From the centre of each face there is one winning line, joining to the opposite face, so there are $3$ 'centre of face' winning lines.
So in total, there are $28 + 18 + 3 = 49$ winning lines.
There are $12$ edges on a cube so there are $12$ winning lines along edges.
There are $6$ faces on a cube, and $4$ winning lines that pass through the middle of each face, so there are $24$ winning lines through the middle of faces.
Finally we need to consider the winning lines that go through the centre cube:
vertex to opposite vertex: $4$
middle of edge to middle of opposite edge: $6$
middle of face to middle of opposite face: $3$
In total, there are $12 + 24 + 4 + 6 + 3 = 49$ winning lines.
On a plane there are $8$ winning lines.
In the cube, there are $3$ horizontal planes, so $8 \times 3 = 24$ winning lines.
There are also $3$ vertical planes going from left to right, but now with only $5$ new winning lines per plane, as the $3$ horizontal lines have already been counted. So $5 \times 3 = 15$ winning lines.
On the $3$ vertical planes going from front to back, we now only have $2$ new (diagonal) winning lines per plane. So $2 \times 3 = 6$ winning lines.
Finally, there are also diagonal planes to consider. There are $4$ winning lines going from corner to diagonally opposite corner.
In total, there are $24 + 15 + 6 + 4 = 49$ winning lines.
Try to make sense of each method.
Now, try to adapt each method to work out the number of winning lines in a $4 \times 4 \times 4$ cube.
Can you adapt the methods to give a general formula for any size cube?
Check that each method gives you the same formula.
You may be interested in the other problems in our Reasoning Geometrically Feature.