My Math Forum Combinatorics with several conditions

 October 21st, 2017, 02:24 AM #1 Newbie   Joined: Oct 2017 From: Czechia Posts: 2 Thanks: 0 Combinatorics with several conditions Hello. I have a math problem that I need to solve, however I'm really stuck and also kind of depressed now. We have 30 balls: 6 red, 7 white, 9 green and 8 blue. We want to separate them to 4 different boxes. • First box can only contain 13 or less balls, which can be either red or white. Number of red balls and white balls in this box must be equal. • Second box can contain 13 or less balls, which can be either green or blue. There must be at least one ball from each of these colors. • Third box can contain 10 or less balls, which can be either blue or red. However there can only be be balls from one of these colors. Not both. • Fourth box can contain 8 or less balls, which can be either white or green. This box can’t stay empty. If the condition doesn't say otherwise, boxes can remain empty as well. I'm supposed to determine the number of possible solutions AND prove that my solution is correct. Unfortunately, creating a tree or creating a computer program to prove my solution is not allowed. It's supposed to be as simple and exact as possible. I think the correct way is to separate conditions to smaller parts (balls go to the box number x or y) and then combine them using intersection and union. Maybe use inclusion and exclusion principle. Do you see it as a viable option? Can you please show me how? I spent too much time on this already to just let it pass. Last edited by skipjack; October 21st, 2017 at 03:33 AM.

 Tags combinatorics, conditions, proof

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