Upper bound for different sums having the same value Let $\Lambda :=\{\lambda_1, \dots, \lambda_n\}$ be a set of $n$ distinct real numbers.
For a given $p \in \mathbb N$, consider further the set
$$I_p := \{  \{i_1, i_2, \dots, i_p\} : i_j \in \{1, \dots,n\} \text{ for all } j=1, \dots, p\}.$$
It is not hard to see that $|I_p| = \binom{n+p-1}{p}$. For example, if $n=3$ and $p=2$, then
$$ I_2 = \{ \{1,1\}, \{1,2\}, \{1,3\}, \{2,2\}, \{2,3\}, \{3,3\} \}.$$
Now for a given $p \in \mathbb N$ we write down all $\binom{n+p-1}{p}$ sums
$$\lambda_{i_1} + \lambda_{i_2} + \dots + \lambda_{i_p}. $$
Clearly it can happen that "different" sums have the same value, e.g. if
$$\Lambda = \{0,1,2\}$$
and $p=2$, then we get the sums $0, 1, 2, 2, 3, 4$ given by considering the elements of $I_2$.
My question is the following: Is there in general an upper bound on how many "different" sums there are that lead to one particular value?
Another interesting question would be the following: Given a fixed upper bound $N^*$, are there conditions on $\Lambda$ such that for all $p \in \mathbb N$, at most $N^*$ different $p$'th order sums yield the same value. For $N^* = 1$ one gets the condition that $\lambda_1, \dots, \lambda_n$ need to linearly independent over $\mathbb Q$.
 A: I answer the first question in Corollary 4.1
of my paper at http://math.mit.edu/~rstan/pubs/pubfiles/42.pdf. The
maximum number of sums that have the same value is the middle
coefficient of the $q$-binomial coefficient $\left[ n+p-1\atop p
\right]$. This is achieved by taking $\Lambda=\{ 1,2,\dots,n\}$ and
the sum to be $\lfloor \frac 12 p(n+1)\rfloor$.  A more elementary
proof follows from the work of Proctor and is given in my book
Algebraic Combinatorics. 
Addendum. For fixed $p$, it follows from http://math.mit.edu/~rstan/papers/qbc.pdf that the middle coefficient of  $\left[ n+p-1\atop p\right]$ is asymptotic to 
   $$ \frac{A(p-1,\lceil p/2\rceil)n^{p-1}}{(p-1)!p!} $$
as $n\to \infty$, where $A(p-1,\lceil p/2\rceil)$ denotes an Eulerian number.
A: A trivial upper bound is ${n+p-1} \choose p$.  Somewhat less trivial is ${{n+p - 1} \choose p}/(p+1)$, which can be obtained as follows.
Suppose we put  the elements in increasing order.  Consider the "stars and bars" representation of $I_p$ (where if there are bars at positions
$1 \le b_1 < b_2 < \ldots < b_{n-1} \le n+p-1$, there are $b_1 - 1$
of $\lambda_1$, $b_2 - b_1$ of $\lambda_2$, ..., $b_{n-1} - b_{n-2}$ of $\lambda_{n-1}$ and $n+p-1 - b_{n-1}$ of $\lambda_n$). 
Suppose we put bars in vacant positions one-by-one.  Each of the $p+1$ possible positions for 
the last bar to be placed produces a different sum (each time you move it to the next available position to the right, you substitute a smaller $\lambda$ in place of a larger one and decrease the sum)).  So the probability of getting a particular sum is at most $1/(p+1)$.
This bound is actually attained in your example with $n = 3$ and $p = 2$.
