17. If 6 colours are available, in how many different ways can one paint each face of a cube with one colour so that adjacent faces are painted in different colours? (Two colourings are regarded to be the same if they are identical upon suitable rotation.)
(2 marks)現有 6 種顏色可供選擇, 並要把一個正方體的每個面塗上一種顏色, 使得相鄰面的顏色不同, 問有多少種塗色方法? (若經適當旋轉後顏色相同, 則兩種塗色方法視爲相同。)
(2 分)
17. 230
17. Note that at least three colours have to be used. So we have four possibilities.
- If exactly three colours are used, there are $C_{3}^{6}=20$ ways to choose colours. Then each pair of opposite faces must be painted in the one colour, and there is only one way of colouring (up to rotation). Hence there are 20 colourings in this case.
- If exactly four colours are used, there are $C_{4}^{6}=15$ ways to choose colours. There must be exactly one pair of opposite faces with different colours, and this pair can be coloured in $C_{2}^{4}=6$ ways. After that the colours of the other four faces are fixed (up to rotation) as each pair of opposite faces must receive one colour. Hence there are $15 \times 6=90$ colourings in this case.
- If exactly five colours are used, there are $C_{5}^{6}=6$ ways to choose colours. There must be exactly one pair of opposite faces with the same colour, and there are 5 choices of colours for this pair (say, top and bottom faces). After that, there are four faces in a ring shape to
be assigned the remaining four colours. We only have to split the four colours into two pairs so that each pair of colours belongs to a pair of adjacent faces. There are 3 ways to do so (just fix one colour and consider which colour to pair up). Hence there are $6 \times 5 \times 3=90$ colourings in this case.
- If all six colours are used, fix any colour and assume (rotate suitably if necessary) that it is for the bottom face. Then there are 5 choices for the colour of the top face. There are now 6 ways to assign the remaining four colours to the four lateral faces. (Not 3 ways as in the previous case, because the top and bottom faces now have different colours, so switching the colours of a pair of opposite lateral faces results in a different colouring.) Hence there are $5 \times 6=30$ colourings in this case.
Combining the four cases, the answer is $20+90+90+30=230$.