A cube with all six faces colored is cut into 64 cubical blocks of the same size which are thoroughly mixed. Find the probability that the 2 randomly chosen blocks have 2 colored faces each?
With some algebra, I found out that each edge is divided into $4$ equal parts. (Since the number of blocks given is $64$).
Now, for the blocks two have exactly 2 colored faces, they must be present at the edges excluding the corners. So, there are a total of $24$ such blocks present.
The total possibility is $\binom{64}{2}$.
So, the probability comes out to be -
$$\frac{24}{\binom{64}{2}}$$
But my book says my answer is wrong.
Any help would be appreciated.
Number of blocks in a cube $=64$
We can choose $2$ random blocks in $\dbinom{64}{2}$ ways
Number of edges $=12\times2=24$
We can choose the edge blocks in $\dbinom{24}{2}$ ways
The probability that the $2$ randomly chosen blocks have $2$ colored faces each is $\dfrac{\dbinom{24}{2}}{\dbinom{64}{2}}=\dfrac{23}{168}$