Bound the number of elements of a set.

Question

Let $A $ a set of real numbers $|A| \lt \infty $.

Let $n \in N $, denote by $A(n) $ the following set: $$A(n) :=\left\{ \sum_{i=1}^{n}a_i ,a_i\in A \right\} $$

prove that : $$|A(n)| \le {{n+|A|-1} \choose {|A|-1}} $$

don't see a way to bound $A(n)$ from above by that number.

Look at the balls and boxes problem, you might find your answer here: Consider every element of $A$ as a box and every number as a ball. Since the ordering of your sum doesn't matter, your balls are indistinguishable. — ctst, Nov 10 '16 at 19:44
See here: [http://math.stackexchange.com/questions/618491/distribute-n-identical-objects-into-r-distinct-groups/618498#618498] — ctst, Nov 10 '16 at 19:45

score 1 · Accepted Answer · answered Nov 10 '16 at 19:46

1

HINT: The binomial coefficient

$$\binom{n+|A|-1}{|A|-1}$$

is the number of solutions in non-negative integers to the equation

$$x_1+x_2+\ldots+x_{|A|}=n\;.$$

Think of the members of $A$ as being numbered from $1$ through $|A|$, and think of $x_k$ as the number of times that you use member number $k$ in forming ... what?

answered Nov 10 '16 at 19:46

Brian M. Scott

616,228

in forming a sum ? – Nov 10 '16 at 20:04
@Liad: Yes. If you use $x_k$ copies of member number $k$ of $A$ in forming a sum of $n$ members, the binomial coefficient gives the number of ways to form such a sum. Some of those ways might yield the same sum, which is why this is just an upper bound for $|A(n)|$. – Brian M. Scott Nov 10 '16 at 20:07
@Liad: You’re welcome! – Brian M. Scott Nov 10 '16 at 20:13

Bound the number of elements of a set.

1 Answers1