Prove that in a 10-element subset of $\{1, \dots , 50 \} $ there exist 2 $5$-element subsets such that the sums of all their elements are equal.

Question

We have a set $$S = \{1, \dots, 50 \}$$ And a given subset $M$ of $S$ containing $10$ elements. We need to prove that there are two $5$-element subsets of $M$ whose elements sum up to the same value.
It looks like a Pigeon Hole Principle to me, therefore my strategy is to prove that in $M$ there are two pairs of numbers that sum up to the same value, namely: $$(\exists a,b,c,d)(a,b,c,d \in M \land a+b = c+d)$$ To do this, I would like to divide all pairs of numbers in $M$ according to their sum. However, I am stuck since I don't know how to compute the number of sums two numbers (unknown!) can generate.
Is my strattegy going to work or it is a dead end?

Answer 1

Note: the header question does not match the question in the body. I answered the one in the header.

There are $$\binom {10}5=252$$

subsets of your $10$ element set with $5$ elements.

The least possible sum is $1+2+3+4+5=15$.

The greatest possible sum is $46+47+48+49+50=240$.

Hence there can't be $252$ distinct sums.

Answer 2

OP asked two questions: one in the header and one in the body.

Header-question. Prove that in a 10-element subset of $S=\{1,\dots,50\}$ there exist two 5-element subsets of $S$ such that the sums of all their elements are equal.

Hint. For any $5$-subset of $S$, the sum of all its elements is less or equal to $50+49+\dots +46=240$. Moreover there are $\binom{10}{5}=252$ $5$-subsets in a 10-element set. Now $252>240$.

Body-question. We have a set $S=\{1,\dots,50\}$. Given subset $M$ of $S$ containing 10 elements. Prove that there are two subsets of $M$ whose elements sum up to the same value.

Hint. If $M$ is any 10-subset of $S$ then the sum of all elements in $M$ is less or equal to $50+49+\dots +41=455$. On the other hand, there are $2^{10}-1=1023$ non-empty subsets in $M$. Now $1023>455$.