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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?

Alessio K
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BodyB
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    But if somone get's $0$ sweets? – nonuser Sep 30 '20 at 18:37
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    When asked to use the Pigeonhole principle, you should at least indicate what the pigeons are, and what the holes are. – Robby the Belgian Sep 30 '20 at 18:39
  • @RobbytheBelgian Okay thanks! So the sweets would be the "pigeons" and the kids would be the boxes/holes – BodyB Sep 30 '20 at 18:42
  • @Aqua Then the kids could all receive separate amounts? As adding (0-19) would only equal 190, meaning that there are enough sweets for each kid to receive a different amount. – BodyB Sep 30 '20 at 18:43
  • @BodyB, no, that is not sufficient. That would just tell us that there is at least one kid who gets 10 sweets. We'll have to be more clever than that. – Robby the Belgian Sep 30 '20 at 18:43

3 Answers3

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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.

José Carlos Santos
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  • Yeah that's what I meant sorry, just in the example I gave where the children receive the sweets in that order then the final kid wouldn't receive 20 but 10. I understand that there are multiple ways the sweets could be distributed but at least 2 children will receive the same amount. – BodyB Sep 30 '20 at 18:40
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Solution is not correct. You may have this distribution: $$0,1,2,3,4....,18,29$$ of sweets.

nonuser
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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).

Nuclear Hoagie
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  • ooo that makes a lot of sense, I was confused by the formula because I assumed the pigeons were the sweets and the holes were the kids. – BodyB Sep 30 '20 at 18:50
  • @BodyB Right, if you set it up that way, you'd be looking for one pigeon (number of sweets) in multiple holes (kids), which doesn't make much sense. Pigeons are always distinct from one another, but the same hole can be used multiple times. If two items can be non-distinct from one another in some aspect (here, number of sweets), then that aspect will be your holes. – Nuclear Hoagie Sep 30 '20 at 18:55
  • Thank you so much for explaining that! – BodyB Sep 30 '20 at 18:57