Pingeonhole Theory, $200$ sweets, $20$ kids. Prove that at least $2$ kids receive the same number of sweets.

Question

I understand the following:

Assuming that every kid gets at least 1 sweet, then $(1+2+...19+20) = 210 > 200$, which is greater than the $200$ sweets stated. Meaning that the kid who received the $20$ sweets would instead receive $10$ which is the same as another kid.

Would this be enough proof if the question states "Use the pigeonhole principle" to answer the question or is there a formula that I should be applying to prove this?

Answer 1

No, it doesn't meant that the kid who received the $20$ sweets would instead receive $10$. It just means that it is not possible that every kid gets at least one sweet and that no two kids get the same amount.

Answer 2

Solution is not correct. You may have this distribution: $$0,1,2,3,4....,18,29$$ of sweets.

Answer 3

Your reasoning is correct (so long as we require that each kid gets at least one sweet), but could perhaps be tied more closely to the pigeonhole principle. In this case, we have 20 pigeons, which are the children, and an unknown number of pigeonholes, which are the number of sweets each kid gets. To have 20 distinct pigeonholes with each one having at least one sweet, you'd require 210 sweets (1+2+3+...+20). Since we do not have that many sweets, there must be less than 20 pigeonholes. With less than 20 pigeonholes and 20 pigeons, two pigeons must share the same hole (i.e. two children must have the same number of sweets).