A gardener plants three maple trees, four oaks, and five birch trees in a row. ..... The question and solution is taken from here. **Question:** A gardener plants three maple trees, four oaks, and five birch trees in a row. He plants them in random order, each arrangement being equally likely. Let $\frac m n$ in lowest terms be the probability that no two birch trees are next to one another. Find $m+n$. **Solution:** The five birch trees must be placed amongst the seven previous trees. We can think of these trees as 7 dividers of 8 slots that the birch trees can go in, making ${}_8C_5$ = 56 different ways to arrange this. There are $^{12}C_5$ = 792 total ways to arrange the twelve trees, so the probability is $56/792$. The answer is $7 + 99 =106$. Can anyone tell me why the total number of ways to arrange the trees is ${12}C_5 = 792$?

It's said in the solution we divide the trees in birch and "non-birch". Let's say we have 12 places where we want to plant the trees and they are numbered. We can freely choose which type of tree we want to plant. And because there are 5 birch trees there are $\binom{12}{5}$ ways to choose how to plant 5 birches in 12 places.

You can do these also with "non-birch" trees and you'll end up with $\binom{12}{7}$, which is completly the same as $\binom{12}{5}$.