Adjacent numbers on a die

Question

Q. Six faces of a cube are numbered randomly $1,2,3,4,5,6$. The probability that faces $1$ & $6$, $6$ & $3$ and $3$ & $1$ will share an edge is __.

My approach: Total ways to write the $6$ numbers on the die $= 6!=720$.

Now, there are $6$ ways to write '$1$'. For each of those $6$ ways, there are $4$ ways to write '$3$' adjacent to '$1$'. And for each of these $6 \times 4$ ways, there are $2$ ways to write '$6$' adjacent to both, '$3$' and '$1$'. So, total favourable ways are $24\times2=48$. Thus, probability $= \dfrac{48}{720} = \dfrac{1}{15}$.

If the correct answer according to the link attached below is $\dfrac25$ please correct me where I am wrong.

Probability regarding die faces

Answer 1

You start correctly. You have 48 ways to place 1,3 and 6.

For each of them, there are 6 ways to place 2,4 and 5 so you need to multiply this by 6.

I spent a long time to detect the error (more than 3 minutes!). The other process (4/5 * 2/4) seems more intuitive. And the risk to forget something is less.

Answer 2

Using Burnside's lemma, there are $30$ ways to construct six side dice using six different numbers $\{1,2,3,4,5,6\}$ such that $$\frac{6!+0+..+0}{24}=30$$

Understanding the lemma is given as homework to reader :)

Now, if these three numbers $1,3,6$ have common sides according to each other, there are $2$ ways to arrange these number on dice.Then, the other three numbers $2,4,5$ can be arranged by $3!$ ways as to any given arrangement of $1,3,6$ with given restriction.Then $$\frac{2 \times 3!}{30}=\frac{12}{30}=\frac{2}{5}$$