[Solved] nCr mod 10^9 + 7 for n
n is large while r is small, you are better off compute nCr by n(n-1)…(n-r+1)/(1*2*…*r) You may need to find multiplicate inverse of 1, 2, … r mod 10^9+7 solved nCr mod 10^9 + 7 for n
n is large while r is small, you are better off compute nCr by n(n-1)…(n-r+1)/(1*2*…*r) You may need to find multiplicate inverse of 1, 2, … r mod 10^9+7 solved nCr mod 10^9 + 7 for n