tag:blogger.com,1999:blog-15628310.post7328759106865442602..comments2020-05-15T09:19:15.838-04:00Comments on Question of the day: How Many Ways Can You Get to 100?Anonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-15628310.post-31560043594838480562009-11-18T08:23:02.290-05:002009-11-18T08:23:02.290-05:00"By removing one stone from each pile, this i..."By removing one stone from each pile, this is the number of ways you can arrange m-k identical stones into k (possibly empty) piles."<br /><br />Hey there, this is a different annonymous user but when you say "possibly empty" is that erroneous considering you are leaving 0 out of the question. (Strictly positive integers)Augiehttp://www.umd.edunoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-31595613374385202822009-11-11T15:33:06.797-05:002009-11-11T15:33:06.797-05:00Hi anonymous, I'm not sure I understand your ...Hi anonymous, I'm not sure I understand your question, so forgive me. But if each of the x, y and z need to be strictly positive integers, then 0 is out and so are decimals.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-63573072179028255232009-11-11T14:01:30.480-05:002009-11-11T14:01:30.480-05:00well technically o is not positive or negative so ...well technically o is not positive or negative so u can not figure one of the numbers as zero and do decimals count? because if so then the answer would b infiniteAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-85864511721356591122007-09-28T08:59:00.000-04:002007-09-28T08:59:00.000-04:00The answer is 4851 ways. It helps if you think o...The answer is 4851 ways. It helps if you think of putting 100 stones into three piles. <BR/><BR/>For a technical answer, read on:<BR/><BR/>By removing one stone from each pile, this is the number of ways you can arrange m-k identical stones into k (possibly empty) piles.<BR/><BR/>Now, view the k piles as a numbered set .<BR/><BR/>Write on each stone the number of a chosen pile and order the stones accordingly.<BR/><BR/>The numbered stones constitute a combination with repetition of k elements (the numbers) choose m-k (the stones). The number of combinations can be calculated as:<BR/> (m-1)!<BR/>---------<BR/>(m-k)!(k-1)!<BR/>In our case, m = 100 and k = 3, so we have 99!/97!*2! = 4851<BR/><BR/>Although, I could be wrong. ;-)Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-46917086660791483882007-09-27T22:06:00.000-04:002007-09-27T22:06:00.000-04:00my unofficial answer will be 5,151 combinations. H...my unofficial answer will be 5,151 combinations. Here's my thinking:<BR/><BR/>where x=0, y+z=100 | 101 combinations<BR/><BR/>where x=1, y+z=99 | 100 combinations<BR/><BR/>where x=2, y+z=98 | 99 combinations<BR/><BR/>therefore, x=n, y+z=101-n | 101-n combinations. <BR/><BR/>101<BR/>sum(x) = (101+1)(101/2)= 5151<BR/>x=0<BR/><BR/>I think that (but dont have the time to check) covers all the combinations because for each value of x, every possible value of y+z is calculated, therefore for every value of y and z, x+y and x+z = 100 respectively. <BR/><BR/>please correct me if my logic is wrong.Anonymousnoreply@blogger.com