A gardener plants 7 maple trees and 5 birch trees in a row. You may assume that all trees are distinguishable from one another. She plants them in a random order, each arrangement being equally likely. Let m/n, in lowest terms, be the probability that no two birch trees are next to one another. Find m+n.
I would have explained it a little bit differently:
I like this approach - it gives C(8,5) for the numerator. That's the number of ways you can arrange the trees so no two birches are next to each other.
My explanation is: for the denominator you need the total number of ways the trees can be arranged. Consider the birches - you have 5 of them and they can go into any of 12 spots, so the number of possible combinations is C(12,5). Or you could consider the 7 maples - they could be positioned in any of C12,7) ways, which is the same thing. So the probability of getting no two birches together is C(8,5)/C(12,5).
I'm not saying my explanation is necessarily better, but for me at least I think it's a bit more intuitive.
Hello, wannabebrainiac!
A gardener plants 7 maple trees and 5 birch trees in a row.
You may assume that all trees are distinguishable from one another.
She plants them in a random order, each arrangement being equally likely.
Let , in lowest terms, be the probability that no two birch trees are next to one another.
Find
There are possible arrangements of the trees.
Place the maples in a row, inserting a space before, after and between them.
. .
There are arrangments of the seven maple trees.
Select five of the eight spaces and insert the birch trees.
. . There are: . ways.
Hence, there are: . ways in which the birches are not adjacent.
Therefore: .
. . and
This is the answer that Plato got.
I just approached the problem slightly differently.