The concept of a tree of life was popularised by Charles Darwin through his theory of evolution through natural selection. The idea is that all species evolved from a common ancestor. Evolution means that at certain points in time a species might split into two new species, which then continue to evolve on their own, perhaps splitting again at some point in the future. When you draw a picture of this you do indeed get something looking a bit like a tree (upside down in the picture below), with today's species at the ends of the branches. The problem is that given a number of species, there is more than one tree that potentially describes their relationships. For example, given three species, there are three potential trees:

So here's a question: given $n$ species, how many different possible trees of life are there? We can answer this by doing a bit of counting. First of all, let's be clear about what kind of object we are counting. We'll assume that a tree of life looks something like the picture below. There is a root which represents the common ancestor. We assume that species can split into at most two (not three or more) other species at a time. Such trees are essentially what mathematicians call rooted binary trees. The tips of the branches, which represent the species alive today, are called the leaves of the tree. A line segment in the tree that connects two neighbouring branching points, or a branching point and a leaf, is called an edge. It's not too hard to convince yourself that a rooted binary tree with $n$ leaves has a total of $2n-1$ edges. You can do this by playing around with a few of these trees, or, much better, by using a proof by induction. A binary rooted tree with $n=2$ leaves obviously has $3$ edges. And since $2n-1$ equals $3$ for $n=2$ this fits the bill. Now suppose the claim holds for any rooted binary tree with $n-1$ leaves. That is, any rooted binary tree with $n-1$ leaves has $2(n-1)-1 = 2n-3$ edges. Now any rooted binary tree with $n$ leaves can be created from one with $n-1$ leaves by picking an existing edge, dividing it in half by a new branch point and have the new leaf sitting on the end of a new edge emanating from the new branch point. This procedure creates two new edges that weren't there before. So if, by our induction assumption, the tree with $n-1$ leaves had $2n-3$ edges, then the tree with $n$ leaves has $2n-3+2 = 2n-1$ edges. This proves the fact that any rooted binary tree with $n$ leaves has $2n-1$ edges. How does that help us count the number of trees with $n$ leaves? Well, first of all notice that if $N_1$ is the number of trees with $n-1$ leaves, then the number of trees with $n$ leaves is $(2n-3)N_1.$ That's because, as we have already seen above, a tree with $n$ leaves can be made from one with $n-1$ leaves by attaching the new leaf to one of the existing $2(n-1)-1=2n-3$ edges. So for every rooted binary tree with $n-1$ leaves there are $2n-3$ rooted binary trees with $n$ leaves: one for every edge of the first tree. But what is the number $N_1$ of rooted binary trees with $n-1$ leaves? Well, we can make a rooted binary tree with $n-1$ leaves out of one with $n-2$ leaves, just as we made a tree with $n$ leaves out of one with $n-1$ leaves. Now, a rooted binary tree with $n-2$ leaves has $2$ fewer edges than a rooted binary tree with $n-1$ leaves. Since a new leaf is attached to one of the edges, we can therefore make two fewer trees with $n-1$ leaves out of trees with $n-2$ leaves than we can make trees with $n$ leaves out of trees with $n-1$ leaves. Writing $N_2$ for the number of trees with $n-2$ leaves, there therefore are $$N_1=(2n-3-2)N_2 = (2n-5)N_2$$ trees with $n-1$ leaves. Similarly, writing $N_3$ for the number of trees with $n-3$ leaves, we see that $$N_2 = (2n-5-2)N_4 = (2n - 7)N_3.$$ You might be able to see where this is going. Working your way backwards, the number $N_0$ of trees with $n$ leaves is $$N_0 = (2n-3)(2n-5)(2n-7)(2n-9)\times ... \times 3 \times 1.$$ This expression is often written as the double factorial $$(2n-3)!! = (2n-3)(2n-5)(2n-7)(2n-9) \times... \times 3 \times 1.$$ It grows very quickly with $n.$ For $n=3$ there are $$3 \times 1 = 3$$ rooted binary trees. For $n=10$ there are $34,459,425$ rooted binary trees. And for $n=20$ there are $8,200,794,532,637,891,559,375$ rooted binary trees. That's one of the problems facing people who study evolution. Even for a small number of species there is a huge number of potential trees describing their potential relationships.

You can read more about the maths of trees of life in Reconstructing the tree of life and Evolution: It's as real as gravity.