Methods of Election. 225 



majority of the electors, better than each of the candidates 

 in the second group, then the proposed method cannot lead 

 to the election of a candidate of the second group. 



The results which have just been proved are obtained 

 from the above by supposing, first, that the first group 

 contains one candidate, and the second group all the rest ; 

 and second, that the first group contains all but one of the 

 candidates, and the second group the remaining candidate. 



Let the first group consist of the I candidates, A, B, C, &c, 

 and let the second group consist of the m candidates, P, Q, 

 R, &c, and let I + m = n, so that n is the whole number 

 of candidates. Because each of the candidates A, B, C, &c, 

 is better than each of the candidates P, Q, R, &c, each of 

 the numbers ap, aq, ar, &c. . . . bp, bq, &c. . . . &c, is 

 greater than N. Now the scores of A, B, C, D, &c, at the 

 first scrutiny are respectively 



* ab -f etc -f ad + $ c -\- ap -\- aq -\- ar -f fyc. 



ba * -f be + bd + &c + bp + bq + br -f fyc. 



ca + cb * .{- cd + fyc -f cp + cq -f cr -f fyc. 



da -f db + dc # + & c + dp + dq + dr + &c. 



fyc. Sec. ^*c. fyc. 



If we add together all these numbers, we shall get the 

 sum of the scores of A, B, C, D, &c. Now the numbers in 

 the first I columns can be arranged in pairs, such as ab, ba, 

 and ab + ba = 2N, and then are \ I (l — 1), of these pairs ; 

 thus, the sum of the first I columns is ~Nl(l — 1). Again, 

 the numbers in the last m columns are each greater than N, 

 and there are Im of these numbers ; thus, the sum of the 

 last m columns is greater than ~Nlm. Thus, the sum of all 

 the numbers is greater than ~Nl(l — 1) + N£m ; that is, 

 than N£(£ + m — 1) ; that is, greater than Nl(n — 1). 

 Thus the sum of the scores of the I candidates of the first 

 group is greater than "Nl(n — 1). Hence the average score 

 of the candidates of the first group is greater than 

 N(w — 1). Hence the candidates of the first group cannot 

 all be rejected at the first scrutiny. By the same reasoning 

 it follows that those of the first group who survive cannot 

 all be rejected at the second scrutiny ; and so on. Thus 

 some candidate of the first group must win on the proposed 

 method ; or, in other words, no candidate of the second 

 group can be elected. 



If the candidates can be divided into two groups in the 

 manner just indicated, it is quite clear that no candidate 

 of the second group ought to win. At the same time, 



R 



