Consider seven integers selected from the first 10 positive integers. Rank these steps to their corresponding step numbers to show that if seven integers are selected from the first 10 positive integers, there must be at least two pairs of these integers with the sum 11. a. Group the first ten positive integers into five subsets of two integers each, each subset adding up to 11.
b. Hence, pairs of integers from groups are obtained. In each case, these pairs of integers have a sum of Il, as desired.
c. Select seven integers from these sets. By the pigeonhole principle at least four of them come from the same subset
d. Leave the two integers in the same subset (group). Then there are five more integers and four groups. Again, by the pigeonhole principle, there are at
least two integers in the same group.
e. Select seven integers from these sets. By the pigeonhole principle at least two of them come from the same subset

Leave the two integers in the same subset (group). Then there are six more integers and five groups. Again, by the pigeonhole principle, there are at least two integers in the same group.