Assume that I have 3 maple trees, 4 oaks and 5 birch trees in a row and I plant them in random order, each arrangement being equally likely. What is the probability that no 2 birch trees are planted next to each other.
I tried to keep 5 birch trees and then I have 6 places to keep the remaining trees and I have 5 trees remaining. So I thought the number of places to plant the trees will be given by the number of positive integral solutions of the equation $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 7$$
which is ${6}\choose{5}$. This multiplied by $7!/3!4!$ should give me the favourable outcomes. This divided by $12!/3!4!5!$ should give me the required probability right? Is my approach wrong because the answer is not matching
Note that the five birch trees must be placed amongst the seven previous trees. So, we can think of these trees as $5$ dividers of $8$ slots that the birch trees can go in $\dbinom{8}{5}$ different ways.
There are $\dbinom{12}{5}$ ways to arrange the $12$ trees.
So the probability is $\dfrac{\dbinom{8}{5}}{\dbinom{12}{5}}$