Zer0's Conundrum #1 has been solved. No additional prizes will be given out. Stay tuned for future conundrums... Welcome to... Zer0's Conundrum #1 So, in Zer0's conundrum, Zer0 will post a problem (usually math or computer science related) and you will have to solve it! First user who solves the problem will win a prize! If no one gets it within a week, there will be no winner for that week. Post your answer here! Rules One entry per person per contest You MUST provide justification (proofs, etc) along with your answer otherwise it WILL NOT count! No collaboration (I don't understand why you would want to...) No looking up the answer or asking someone for the answer No computer programs unless otherwise specified This Week's Prize... 15 forum cash! - courtesy of some mod! 50k NP! - courtesy of Dark This Week's Conundrum... How many different ways can one select 2 distinct points from a 5x5 grid of points such that they are not in the same row or column and neither point is both above and to the left of the other? (The order in which the two points are selected does not matter)
Re: Zer0's Conundrum #1 does order matter? like would the pair (x1,y1) (x2,y2) be different than (x2,y2) (x1,y1)? if order does matter - 400 (25 points * 16 pairs for each) if order does not matter - 200 25 (25 points * 16 pairs for each / 2 since order doesn't matter)
Re: Zer0's Conundrum #1 184 I don't really know how to explain this well... I basically separated the points into three sets: points without a point that is both above and to the left of the other, points with one point that is both above and to the left of the other or both below and to the right of the other, and points with both a point that is both above and to the left of the other and a point that is both below and to the right of the other. 2*16 = 32 14*15 = 210 9*14 = 126 32+210+126 = 368 368/2 (since order doesn't matter) = 184
Re: Zer0's Conundrum #1 If you don't have a reason, then its incorrect. Please don't post answers without actually thinking through the problem. Otherwise its just spam.
Re: Zer0's Conundrum #1 You can have 2 different points on the 5x with neither of them in the same row or column 222 times, I think. lol The way I came up with my answer was I had drawn a 5x5 then started doing what you asked. First I did all the diagonal combinations, then went row by row, was going to do column by column but then saw the overlapped and then stopped lol Had 444, divided it by 2 since you only wanted the amount of times the 2 points can be on there and not in the same row or column =P How bad did I do? lol ~Parisdirt
Re: Zer0's Conundrum #1 100. Just systematically going through point by point starting in upper left and counting unique pairs gives the following table: 0, 4, 8,12,16 0, 3, 6, 9,12 0, 2, 4, 6, 8 0, 1, 2, 3, 4 0, 0, 0, 0, 0 Do I need a better proof? EDIT: If you don't understand, it's just the number of points below and to the left of each grid point...this is why the left hand column and bottom row are all 0s...
Re: Zer0's Conundrum #1 correct I will go bother some people about your prize. I'll contact you once I get that all sorted out