It's equivalent to (x1-1)+(x2-1)+. . .+(xk-1)=n-k (You need n>=k) yi = xi - 1, so for every 1<=i<=k, we have that yi is non-negative.
Now think about it as having (n-1) objects in a row, and you need to choose k-1 which will be black and the other n-k will be white so, the number of solutions is equal to (n-1)C(k-1). 2010/4/9 GentLeBoY <[email protected]> > no. of solutions to linear equation as > x1+x2+x3+. . .+xk=n , all variables are positive natural numbers > how is it (n-1)C(k-1) plz explain > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]<algogeeks%[email protected]> > . > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- Mario Ynocente Castro Undergraduate Student of System Engineering National University of Engineering, Peru -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
