tag:blogger.com,1999:blog-15628310.post2292208135248115997..comments2019-10-14T19:41:27.230-04:00Comments on Question of the day: Vacation From SchoolUnknownnoreply@blogger.comBlogger3125tag:blogger.com,1999:blog-15628310.post-26671400908937614702008-12-24T07:52:00.000-05:002008-12-24T07:52:00.000-05:00Nice work, both of you. Here's another way to exp...Nice work, both of you. Here's another way to explain it, although oudeis did a great job above.<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/>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). This can be done in<BR/><BR/><BR/>C'(k,m-k) = C(m-1,m-k) = <BR/><BR/>(m - 1)!<BR/>----------------- ways.<BR/>(m - k)!(k - 1)!Mikehttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-48087006404732089222008-12-24T04:50:00.000-05:002008-12-24T04:50:00.000-05:0098+97+96+95+......+3+2+1 = 485198+97+96+95+......+3+2+1 = 4851sadhttps://www.blogger.com/profile/10094105058715623592noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-65461338715704675922008-12-23T16:29:00.000-05:002008-12-23T16:29:00.000-05:00I'll take a crack at it; Let me see:If the pil...I'll take a crack at it; Let me see:<BR/><BR/>If the piles are non-identical (indexed with 1,2,3), then the first pile may have n = between 1 and 98 stones. In each of these cases, the number of stones in the second pile can range from 1 to (100 - (n + 1)), and after that the distribution is uniquely identified. So the total number of combinations is SUM(n=1 to 98)[99-n], or SUM(1,2,3,...,98), which is equal to 4,581.<BR/><BR/>If the piles are identical (so that, for instance, {1,1,98} is the same distribution as {1,98,1}), we can group the combinations above and count the number of such groups, leaving out the "repeated" combinations. There is no {n,n,n} combination, since 100 % 3 <> 0. There are 49 {n,n,m} combinations (one for each number such that m = 100 - 2*n > 0), and each occurs 3 times ([n,n,m],[n,m,n],[m,n,n]). The remaining 4,434 (4,581 - 49*3) combinations are [n,m,l] combinations with no repeated entries, and they occur 6 times each ([n,m,l],[n,l,m],[m,n,l],[m,l,n],[l,m,n],[l,n,m]), for a total of 739 unique combinations. Now, 49 + 739 = 788; so I believe that is our answer?oudeishttps://www.blogger.com/profile/02798986999913350014noreply@blogger.com