value of $k$ in cube painting with $6$ different color

Question

Let $k$ be the number of distinct ways of painting $6$ sides of a cube with $6$ distinct colors, where you would regard two cube coloring the same If one can be obtained from the other by rotation, Then $k$ is

what i try

Number of distinct ways of painting a cube with $6$ distinct color is $\displaystyle 6\times 5\times 3!=180$

I have color Top side, can be done as $6$

and opposite face bottom side with $5$ color Now arranging $4$ color in $4$ sides as $(4-1)!$

But my answer is wrong, Help me please

Answer 1

First, you need to know/assume that only 6 colours are available. Call them $1,2,3,4,5,6$. You can always rotate the cube so that colour 6 is on the top face.

Now there are 5 choices for the colour of the opposite face.

Now you can rotate the cube about a vertical axis so that the highest remaining number is in some given position. There are then 6 ways of painting the remaining three faces. So 30 ways in all.