Very nice solution. Don On Wednesday, April 9, 2014 12:55:43 AM UTC-4, Dave wrote: > > What you want to do is automate the process of cancelling common factors > that you would do if you were calculating nCr by hand. Fill an array a[] > with the numerator terms, n, n-1, n-2, ..., n-r+1, and a second array b[] > with the denominator terms, 1, 2, 3, ..., r. Inside a double loop with j > and i going from 0 to r-1, compute the gcd of a[i] and b[j]. Divide both > a[i] and b[j] by the gcd. When the loops finish, you will have cancelled > the common factors from all of the terms, and in fact, all of the > denominator terms will be 1. In a final loop, compute the product of the > numerator terms. There are some obvious optimizations. > > Dave > > On Tuesday, April 8, 2014 1:06:47 PM UTC-5, kumar raja wrote: > >> Hi all, >> >> I know the way to calculate value of C(n,r) using pascals identity. But i >> have a different requirement. >> >> C(n,r) = n (n-1) (n-2) ... (n-r+1) / r! >> >> Suppose the numerator is such that it cant fit into existing data type. >> >> So i remember there is a way to strike off the common factors in >> numerator and denominator then calculate the result of it. But the logic >> is not striking to me. Googled but could not find much. So please share the >> clever way to calculate the above value w/o actually computing the >> factorials and then making division. >> >> My actual requirement is to calculate the value of >> >> (n1+n2+....+nk)! / (n1!.n2!.n3!.....nk!) . >> >> >> Regards, >> Kumar Raja. >> >> >>
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