Basically you are looking to count the number of unique binary expression trees where the nodes are summations and the leaves are x1 to xn. Let's call this number P(n).
You can choose any of the n-1 summations as the root node. Let's choose the k'th summation. There are k variables to the left of the root, so you can organize the left subtree in P(k) ways. There are n-k variables to the right, so the right subtree can be organized in P(n-k) ways. So, in total you can organize the tree in P(k)P(n-k) different ways.
Since you could choose k freely from 1 to n-1, the total number with which you can organize the expression tree with n leaves is:
P(n) = P(1)P(n-1) + P(2)P(n-2) + ··· + P(n-2)P(2) + P(n-1)P(1)
As noted by @DSM in the comments this recurrence relation generates the Catalan numbers. There is a closed form solution, but it's a little tricky to derive from the recurrence formula. You can find several proofs of the formula in the Wikipedia article for Catalan numbers. The solution is:
P(n) = C(2n,n)/(n+1) where C(n,k) = n! / (k!(n-k)!)