(nCr + nC(r+1) + nC(r+2) + .... + nCn)/2^n
value of this is atmost 1,but for large n(10^5) nCr value overflows. can any1 give me an efficent approach to solve this(r<=n). thanks.
how to calculate the value of following question without overflow?
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$\begingroup$
probability
combinatorics
combinations
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0$nCr = {n \choose r}$ ? – 2017-02-13
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0http://math.stackexchange.com/questions/2135431/how-to-calculate-the-summation-sum-p-kn-binomnp-2n-quickly – 2017-02-13
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0To avoid overflow when working with ratios of factorials and powers you can make us of the Log Gamma Function which can be efficiently numerically calculated. – 2017-02-14