Pigeonhole again

Question

There are $10$ bird cages with the maximum of $5$ birds inside it. How many birds should be prepared so that I can be sure that there are $3$ cages with $2$ birds inside it?

My answer :

I fitted all the cages with $1$ bird, that means $10$ and I filled the other $3$ cages with $1$ more bird. That makes it $13$ And everyone said it isn't $13$, to be sure I need $19$ birds. Can someone make me understand?

Thanks

I don´t really understand the question. Are there $10$ cages filled with $5$ birds each and we have to remove birds to make the condition that "there are $3$ cages with $2$ birds inside" or we start with empty cages and fill them with birds making sure that the condition holds? Reading your answer makes me think the latter but in that case I don´t understand why $19$ is the solution. — Vinyl_cape_jawa, Feb 26 '19 at 15:49

score 3 · Answer 1 · answered Feb 26 '19 at 14:54

3

You can fill all 10 cages with 1 bird (as you did), but then fill two cages to capacity (so another 8 birds) without having 3 cages with 2 birds.

answered Feb 26 '19 at 14:54

Aaron

24,207

This adds up to $18$ or am I missing something? – Vinyl_cape_jawa Feb 26 '19 at 15:49
@Vinyl_coat_jawa That's 18 that you can place without satisfying the problem. If you place 19 (the number mentioned in the problem), at least 3 of the cages will have at least 2 birds each. – Aaron Feb 26 '19 at 16:03
Ahhh! I am with you now, this wasn´t obvious from the text of the question. Thank you for elaborating! – Vinyl_cape_jawa Feb 26 '19 at 16:06

Pigeonhole again

1 Answers1