# Choosing $7$ numbers from $[1,2,…,11]$ will give us $2$ that have sum $12$.

Choosing $7$ numbers from $[1,2,...,11]$ will give us $2$ that have sum $12$. I tried:

There are only $5$ pairings possible:

$(7,5),(8,4),(9,3),(10,2),(11,1)$

Suppose I pick $6$, and then not to be a pair I will need to draw exactly 1 number from each of the above pairs. Suppose I picked, $7,8,9,10,11$. I will use up $6$ numbers. Thus any number that is left belongs to some pair.

How can I put this into mathematical terms and prove it?

Prove that for any set of $7$ numbers chosen from the set $\{1, 2, 3, ..., 11\},$ there exists at least one pair of two numbers that sum to $12.$
We can formally prove this by contradiction. Imagine that we can find a set of $7$ numbers with no two elements with a sum of $12.$ We have $6$ boxes - one of which can hold up to $1$ number. Then our proposition is that each of $b_{1}, b_{2}, b_{3}, ..., b_{6}$ is less than or equal to $1.$ In other words, $b_{n} \le 1.$ Then $$\sum_{n = 1}^{6} b_{n} \le 6.$$
• I meant that one box can hold up to 1 number. It can only hold $6.$ Because $6 + 6 = 12,$ and there is only 1 $6$ in the set. – K. Jiang Mar 10 '16 at 0:27