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?**

Click below to see four different methods for working out the number of winning lines in a $3 \times 3 \times 3$ cube.

**James’ method**

Winning lines can either be:
**non-diagonal winning lines** first:

There are $9$ from front to back.

There are $9$ from left to right.

There are $9$ from top to bottom.

Considering the**diagonal winning lines:**

On each layer there are $2$ diagonal winning lines so:

There are $6$ from front to back.

There are $6$ from top to bottom.

There are $6$ from left to right.

There are $4$ lines from a vertex to a diagonally opposite vertex.

In total, there are $27+18+4=49$ winning lines.

- Diagonal
- Not diagonal

There are $9$ from front to back.

There are $9$ from left to right.

There are $9$ from top to bottom.

Considering the

On each layer there are $2$ diagonal winning lines so:

There are $6$ from front to back.

There are $6$ from top to bottom.

There are $6$ from left to right.

There are $4$ lines from a vertex to a diagonally opposite vertex.

In total, there are $27+18+4=49$ winning lines.

**Alison’s method**

There are $3$ possible places where a line can start:

- at a vertex
- at the middle of an edge
- in the centre of a face

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.

**Grae's method**

The winning lines may be counted by looking at lines:

- in each horizontal plane
- in each vertical plane from left to right
- in each vertical plane from front to back
- in the diagonal planes

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.

**Caroline’s method**

All winning lines must pass either:

- along an edge of the cube
- through the middle of a face
- through the centre of the cube

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.

**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.