Show that there are two 5-element subsets

Question

I have such question and I don not know how to handle it.

Task

$A\subset \left \{ 1,..., 50 \right \}, |A|=10$ . Show that there are two 5-element subsets of A with the same sum of elements.

Answer 1

The maximal sum a 5 element subset of $A$ could possible have, is $50+49+48+47+46 = 240$, the minimal sum is $1+2+3+4+5=15$ so there are at most $226$ possible sums. Now there are $\binom{10}{5} = 252$ different 5-element subsets of $A$. By the pigeon hole principle there must be two such subset that have the same sum.

Answer 2

How many such subsets are there?

How many possible values do you have for the sum of the elements?

Answer 3

Use the pigeon hole principle. There are ${10\choose{5}} = 252$ subsets of $A$ of size $5$. Each of these has some sum that lies in the set $S=\{15, 16,\dots 240\}$. The limits for this set $S$ come from the fact that the minimal sum is $1+2+3+4+5=15$ and the maximal sum $46+47+48+49+50=240$. So there are only $226$ possible values for the sum. That means there must be two subsets of $A$ of size $5$ that have the same sum.