# Prove that there are at least three balls which lie in the same box, have the same colour and are of the same size depending upon following condition.

Two boxes contain between them 65 balls of several different sizes. Each ball is white, black, red, or yellow. If you take any five balls of the same colour, at least two of them will always be of the same size (radius). Prove that there are at least three balls which lie in the same box, have the same colour and are of the same size.

My approach:-

Ans:- Making repeated use of pigeon-hole- principle (PHP)., there are 65 balls and 2 boxes, one of these boxes must contain at least $$\left[\frac{65}{2}\right]+1=33$$ balls. Consider that box, now we have four colours (white, black, red, yellow) and hence there must be at least $$\left(\frac{33}{4}\right)+1=9$$ balls of the same colour.

What to do next? How can i proof atleast three balls are of the same size.

## 2 Answers

You're almost there.

Since for any five balls of the same color, at least two of them will be of the same size (radius), it follows that for a given color there are at most $$4$$ distinct sizes.

Can you finish it?

• But as there should be atleast 2 balls of same size,henceworth there must at most 3 different size of balls among 9 balls. You are saying "there are at most 4 distinct sizes" Suppose one choose those 4 balls and remaining 1 from others' same size then it will violate the condition which is given in the problem.I am getting confused – user791682 Jun 23 '20 at 6:19
• For a given color, since there are at most $4$ distinct sizes, then to have at most $2$ of each size, you can have at most $4{\,\cdot\,}2=8$ balls. – quasi Jun 23 '20 at 6:24
• great! got it.thanks – user791682 Jun 23 '20 at 6:26

Hint: From one of conditions it follows that there are no more than 4 different sizes in the group of 9 balls.