tag:blogger.com,1999:blog-15628310.post6534849696364234428..comments2022-07-26T20:48:08.865-04:00Comments on Question of the day: Another Counting ProblemAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-15628310.post-811786599512760252010-12-04T03:48:38.659-05:002010-12-04T03:48:38.659-05:00Ofcourse the answer is gonna be 120 since we would...Ofcourse the answer is gonna be 120 since we would have to find the answer by keeping one object fixed and then finding the possible ways.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-18086050493906474832007-10-02T22:37:00.000-04:002007-10-02T22:37:00.000-04:00Circular permutations. From a placement point of v...Circular permutations. From a placement point of view, ask the first person to sit. Doesn't matter where. Now, the second person has to make a choice, and the third, etc.<BR/>1 x 5 x 4 x 3 x 2 x 1Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-32697495939538964892007-10-01T08:10:00.000-04:002007-10-01T08:10:00.000-04:00I think you've figure it out Abe. It actually was...I think you've figure it out Abe. It actually wasn't the answer I was thinking of, but your answer makes a lot of sense.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-73105354901042164792007-09-28T12:07:00.000-04:002007-09-28T12:07:00.000-04:00Well, let's look at how many ways they can be arra...Well, let's look at how many ways they can be arranged in a linear pattern. This is a classic factorial problem, so the answer to that is 6!<BR/><BR/>However, when you make the pattern circular, for each pattern that you have, the "starting number" is arbitrary. This means that 123456 = 234561 = 345612 = ...<BR/><BR/>So, each pattern with "1" as the starting number takes care of 5 others (1 pattern -> 6 patterns). So, I'd just divide by 6.<BR/><BR/>6!/6 = 5*4*3*2*1 = 120Abehttps://www.blogger.com/profile/04424868492071587450noreply@blogger.com