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In case (c), the x Electors can, as in case (a), supply s Candidates with vxx votes apiece. But each of the (e − x) Electors can only use (m + 1 − s) of his votes: hence the (e − x) Electors can only give (m + 1 − s).(e − x) votes, thus supplying (m + 1 − s) Candidates with (e − x) votes apiece. Hence we must have
| vxs > | e − x; |
| ∴ vx > | se − sx; |
| ∴ x.(s + v) > | se; |
| ∴ x > | ses + v. |
In case (d), the x Electors can, as in case (b), supply s Candidates with x votes apiece. And the (e − x) Electors can, as in case (c), supply (m + 1 − s) Candidates with (e − x) votes apiece. Hence we must have
| x > | e − x; |
| ∴ 2x > | e; |
| ∴ x > | e2. |
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