Finding two integers to equal 26

Question

Show that if any 14 integers are selected from the set $S = \{1,2,3,...,25\}$, then there are at least two whose sum is 26.

Let there be finite sets $A, B, C $. $A=\{ X| X\in\mathcal{P}(S)$ and $|X|=14 \}$ $B=\{ X| X\in\mathcal{P}(S)$ and $|X|=2 \}$ $C=\{ X| X\in\mathcal{P}(S)$ and $|X|=2$ and $x_1+x_2=26, x_1,x_2\in X\}$

Then, $C \subseteq B $. $C$ is nonempty because $\{1, 25\}\in C$. $|A|=\binom {25}{14}= 4457400$ $|B|=\binom {25}{2}=300$ Since $|A| >|B|$ and $|B| >=|C|, |A|>|C|$. Thus, the function $f: A\rightarrow C $ is not injective by the pigeonhole principle. Suppose $A_1=A_2, A_1, A_2 \in A$. Then $f(A_1)\neq f(A_2)$ because f is not injective.

I am kind of stuck on the proof now. If you choose any subset in A, then how do you ensure that the corresponding subset in C contains elements of A? Is this even the right approach to the proof?

Answer 1

You can have at most one number in each of the following 13 boxes. The 14th gives you two in one box, and you have your result.

1,25 | 2,24 | 3,23 | ... | 12,14 | 13

Answer 2

consider sets$$(1,25),(2,24),...,(12,14),(13)$$

by pp we have atleast one pair from the same box hence the sum would be $26$