Question

The number of ways of painting the faces of a cube with six different colours is
[a] 30
[b] 6
[c] 6!
[d] None of these.

Answer 1

Hint: There are two cases of the problem which we need to consider. Case I being when the faces of the cube are distinct. Case II being the Case when the faces of the cube are not distinct. Observe that the first Case is equivalent to the arrangement of 6 people in 6 seats. Use the fact that the arrangement of r people among n people in r places can be done in $^{n}{{P}_{r}}$. Hence determine the number of ways of colouring the cube in Case I. Observe that in Case II, 8 different combinations of Case I are actually only one distinct combination. Hence determine the number of ways of colouring the cube in Case II.

Complete step-by-step answer:
We need to consider two cases to solve the question:
Case I: When all the faces are considered to be distinct, e.g. the faces of a dice.
In this case, we can colour face 1 in 6 different ways, face 2 in 5 different ways(excluding the colour used in face 1), face 3 in 4 different ways, face 4 in 3 different ways, face 5 in 2 different ways and face 6 in one way.
We know from the fundamental principle of counting if the number of ways in which we can do task A is m and the number of ways in which we can do task B is n, then the number of ways in which we can do both the tasks A and B is mn.
Hence by the fundamental principle of counting, the number of ways in which we can colour the cube is $6\times 5\times 4\times 3\times 2\times 1=6!$ ways.
Hence the number of ways of colouring the cube with 6 different colours is 6!.
Hence option [c] is correct.
Case II: In this case the faces of the cube are not distinct, i.e. the opposite faces of the cube are identical.
Since the opposite faces of the cube are identical, if we interchange the colours between them, the net colouring of the cube does not change. Now for every colouring of the cube, the interchange between opposite faces of the cube can be done in $2!\times 2!\times 2!=8$ ways.
Hence every combination of Case II is considered 8 times in Case I
Hence, we have
The number of ways of colouring the cube in Case II is $\dfrac{6!}{8}=90$
Hence option [d] is correct.

Note: [1] For exam purposes, if nothing about the cube is said, we consider the cube to be in Case II and hence option [d] will be correct. There is a common mistake of overcounting in the questions of permutations and combinations which many students make. This can be avoided by making a few combinations using the argument applied in solutions and checking for repetition.