Suppose we wrote $11$ numbers $1,2,3,4,5,6,7,8,9,10,11$ each on different piece of paper and then put all those papers in a box and we shuffle those pieces.
Now how many minimum number of pieces we need to take out from that box to ensure that the number written on any $2$ of the pieces that we are taking out from the box results in difference of $7$ without looking at the numbers that we are picking out from the box..
here the name of the box is $S$. i.e. box=set.
The pairs that would result in difference of $7$ are $(1,8) ,(2,9), (3,10),(4,11)$ only
So we have to calculate the worst case of getting $7$ as difference of any $2$ numbers that we ae picking out.
We pick $1$ then $2$ then $3$ then $4$ then $5$ then $6$ then $7$ uptil here the difference of none of the $2$ numbers that we picked results in $7$
Now whatever number we pick next from the remaining numbers in the set(box) we would definitely get $7$ as the difference.i.e.
if we pick $8$ then we can get $8 -1 =7$
if we pick $9$ then we can get $9-2 =7$
if we pick $10$ then we can get $10-3 =7$
if we pick $11$ then we can get $11-4=7$
$\therefore$ The minimum numbers that we need to pick are $8$ to ensure that we get a difference of $7$.