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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 $7$.

by (3.7k points)
0
How 8 is should b 4 na
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Ohkk got it
0
yes 1,2,3,4,5,6,7,(8 or 9 or 10 or 11)