How many numbers need to be selected to guarantee that at least one pair of these numbers add up to 80? Suppose someone is randomly selecting numbers from the set:
$\{2x-1\ |\ 1\leq x \leq 40\} = \{1, 3, 5,...,75,77,79\}$
without repetition. How many numbers need to be selected to guarantee that at least one pair of these numbers add up to $80$?
The answer came to be $21$. How is this so? 
Please provide clear explanation, if possible. Thanks. 
 A: Group the set into pairs as such: $$\left\{\{1,79\},\{3,77\},\cdots,\{39,41\}\right\}$$
Observe that the pairs all add up to $80$, and there is no other way to add up to $80$.
Therefore, we need to choose randomly such that both numbers of any pair is chosen.
The worst case scenario would be that no pairs have both numbers chosen.
We have $20$ pairs, so the worst case scenario would be that only one number of those $20$ pairs are chosen: $$\{1,3,\cdots,39\}$$
Therefore, we would need to choose $21$ times to guarantee that.
A: The set $S = \{2k - 1 \mid k \in \mathbb{N}, 1 \leq k \leq 40\} = \{1, 3, 5, \ldots, 79\}$ consists of the $40$ positive odd numbers less than $80$. A pair of odd numbers that has sum $80$ has the form $\{m, 80 - m\}$.  The set $S$ contains $20$ such pairs:
$$\{1, 79\}, \{3, 77\}, \{5, 75\}, \ldots, \{39, 41\}$$
To guarantee that we obtain a pair with sum $80$, we must select both numbers from one of these two-element subsets.  The worst case scenario is that as we keep making selections, we choose exactly one number from each of these twenty subsets, so selecting only $20$ numbers from set $S$ does not guarantee that we will obtain a pair that has sum $80$.  On the other hand, if we select $21$ numbers from set $S$, by the Pigeonhole Principle, we must select both numbers from at least one of these $20$ two-element subsets. Consequently, if we select $21$ numbers from set $S$, we will select at least one pair with sum $80$.  
A: The question is based on pegionhole principle.We have to contradict the statement and as maximum we can contradict we get the answer.So let's contradict:-
From the given numbers $80=1+79=3+77=5+75$ & so on......As there are 80 odd numbers ,so the total numbers we have are $40$ out of which $20$ pairs can be made which add upto $80$.
Now the contradiction begins :- pick all the numbers which can not add upto $80$ in pair of two.Pick the beginning $20$ numbers because we have seen that without their pair from the backend they are not making $80$.Now we have got the maximum negativity($20$) because if we take one more number i.e. $41$ we have $39$ for that.So we have to take $21$ numbers to be sure.
If we think of it another way ,let's take any of the $20$ numbers that don't add upto $80$ in pair of two ,the second number doesn't have its pair, the third one also don't have & so on upto $20^{th}$ ,so logically if $(A,B)$ is the pair then we have taken either $ A $ or $B$ of each of the $20$ pairs, thus the next number we select would be the counterpart of any one of the $20$ already selected numbers(if that was $A$ then it would be $B$ or vice versa),because there is no $21^{th}$ pair ,so we have to take $21$ numbers to be sure.
