<s>[INST] Simplify conditional probability tree There are six regular (6-sided) dice being rolled. However, each dice has one side colored gold. The 1st has a gold "1", the 2nd has a gold "2"... and the 6th has a gold "6".
This question is similar to my last question except the desired outcome is rolling one of each number, the probability being:
$$\frac{6!}{6^6} = \frac{720}{6^6}$$
An example of this kind of roll would be "123456" or "352461".
I need to determine the breakdown of the amount of rolls by number of gold sides. Using the method of the accepted answer I could get the number of rolls with 6, 4, and 3 gold sides (exactly 5 gold sides is impossible as the last side must also be gold in order for the roll to contain one of each number).
$$G_6 = \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\times 6! = 1$$
$$G_4 = {6 \choose 4} \times \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{1}{3} \times \frac{1}{2} \times \frac{1}{1}\times 6! = 15$$
$$G_3 = {6 \choose 3} \times \frac{1}{6} \times \frac{1}{5} \times \frac{1}{4} \times \frac{2}{3} \times \frac{1}{2} \times \frac{1}{1}\times 6! = 40$$
Now for $G_2$ the probability of the 4th dice being gold became conditional on the position of the 3rd dice, and the 5th conditional on the 3rd and 4th. I drew a probability tree to find there are 9 positions out of 24 for the last 4 dice not to be the number of their gold side.

And then plug that in to find the answer:
$$G_2 = {6 \choose 2} \times \frac{1}{6} \times \frac{1}{5} \times \frac{9}{24} \times 6! = 135$$
But I am wondering what is the better way to find the answer (without having to draw the tree, as a tree for $G_1$ and $G_0$ would be even larger and more tedious).
 [/INST] There is no easy way to do it with a probability tree. For any $k\in \{0,\dots,6\}$, the number of rolls with $k$ gold faces is equal to the number of permutations of $\{1,\dots,6\}$ with $k$ fixed points, which is in turn equal to $\binom{6}k$ times the number of derangements of a set of size $6-k$. The number of derangements of a set of size $n$ is well known to be
$$
n!\sum_{i=0}^n\frac{(-1)^i}{i!}
$$
which can be found using the principle of inclusion-exclusion.
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