Page:Carroll (1884).djvu/30

 the (e − x) Electors can give v.(e − x) votes, which, divided among (m + 1 − s) Candidates, supply them with $v.(e − x)⁄m + 1 − s$ votes apiece. Hence we must have

In case (b), each of the x Electors can only use s of his v votes, since he can only give one to each Candidate: hence the x Electors can only give sx votes, thus supplying s Candidates with x votes apiece. But the (e − x) Electors can, as in case (a), supply (m + 1 − s) Candidates with $vx⁄s$ votes apiece. Hence we must have