tag:blogger.com,1999:blog-15628310.post3377006002761390263..comments2021-04-22T08:14:35.383-04:00Comments on Question of the day: Connecting the DotsAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-15628310.post-71007799833552037992007-10-15T07:49:00.000-04:002007-10-15T07:49:00.000-04:00To say I'm impressed by these answers would be an ...To say I'm impressed by these answers would be an understatement. As I said before, the answer is well known, but the explanations you gave were thorough.<BR/><BR/>I'm not really sorry the question was vague, though. To see so many possibilities explained and discussed helps me think things through, and I hope it does the same for everyone else who reads them.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-67545720266337254942007-10-14T15:17:00.000-04:002007-10-14T15:17:00.000-04:00Mike, I think what you were meaning is n, non-coli...Mike, I think what you were meaning is n, non-colinear points. In order to draw lines connecting n points that meet that criterion, the number of lines you would need is:<BR/><BR/>n!/(2*(n-2)!)<BR/><BR/>Basically, for each line, you're "choosing" two points, and order does not matter, so it is the general formula for "combinations" (similar to the last post, only order does matter here).<BR/><BR/>Incidentally, for our case, the above formula reduces to:<BR/><BR/>n!/(2*(n-2)!)<BR/> = (n*(n-1)*(n-2)!)/(2*(n-2!)<BR/> = n(n-1)/2<BR/><BR/>which is the same as benvitale's answer (as well as consistent with everyone else's more empirical results).Abehttps://www.blogger.com/profile/04424868492071587450noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-49164886531569858172007-10-14T14:45:00.000-04:002007-10-14T14:45:00.000-04:00The problem as stated is too broad, it needs clari...The problem as stated is too broad, it needs clarification.<BR/>If we on't consider the case of colinear points, then we could say:<BR/><BR/>it is the same as finding the number of pairs from a set of objects.<BR/>Because each pair of dots define a line.<BR/><BR/>For example: <BR/><BR/>If you have 2 points A, B in a space, then you have AB and BA, or 2*1 = 2<BR/>possibilities.<BR/>If you have 3 points A, B, C in the space, then you would have: <BR/>AB, BA, AC, CA, BC, CB or 3*2 = 6 possibilities. <BR/><BR/>You may argue that AB and BA are the same, thus giving 1 possibility <BR/>Similarly, that AB=BA, BC=CB and AC=CA, thus giving you 3 possibilities. <BR/><BR/>So, we could say that the number of pairs is 1/2 the number of<BR/>possibilities.<BR/><BR/>With n=4 points, we have 4*3/2 = 6<BR/><BR/>With n points, n(n-1)/2.<BR/><BR/>But, if we consider that the points could be colinear, then the number of lines are between 1 and n(n-1)/2.BenVitalehttps://www.blogger.com/profile/08252218231924085041noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-34979833468002980902007-10-13T10:35:00.001-04:002007-10-13T10:35:00.001-04:00no, i guess notno, i guess notAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-84903300331079470412007-10-13T10:35:00.000-04:002007-10-13T10:35:00.000-04:00n(sub x)=n(sub x-1)!or: (I dont know if this will ...n(sub x)=n(sub x-1)!<BR/><BR/>or: (I dont know if this will come out right when posted.)<BR/><BR/>n =n !<BR/> x x-1Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-45986866832113549682007-10-12T18:46:00.000-04:002007-10-12T18:46:00.000-04:00Well, not sure how to write this into an equation ...Well, not sure how to write this into an equation (or as an answer) but, if n was:<BR/>1, there would be 0 lines connecting<BR/>2=1<BR/>3=3<BR/>4=6<BR/>5=10<BR/>6=15<BR/>The pattern is shown in the difference between the number of lines (1,2,3,4,5)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-82758879220163947592007-10-12T17:47:00.000-04:002007-10-12T17:47:00.000-04:00Is the answer not "n-1"?Is the answer not "n-1"?Anonymousnoreply@blogger.com