188 



Summation of Polynomial Co-efficients. 



[May, 



a'a + a'b 4- a'c + &c. (to the nth term) = n combinations. 



b'a + b'b + b'c + &c. = n ditto. 



c'a -f- c 'b + cc + & c « =£= w ditto, 



that is, each of the n letters of one set forms n combinations with the 

 letters of the other set : the number of combinations isn x n = n 2 . 

 Since the numeral co-efficient of each combination is 1, n* likewise 

 represents the sum of their numeral co-efficients. 



Next, let there be three sets, each containing n letters, taking two 

 sets at a time, and one letter from each of those two sets at a time. 



r There are only three 



V possible ways of ar- 



J ranging three sets tak- 



3. 2 



en two at a time-; — - 



1. 2 



a+ b -f-c &c. 

 «* + &'-|.c'&c. 



a-\- b + c &c. 



a'-f.6'-f- c'&c. 

 a"+b"+c"kc. 



I 



= 3. 



The first arrangement according to (2) produces n 2 combinations of 

 letters ; the second n* combinations also, and the third n 2 combinations, 

 the total number of combinations = 3 /i% and for the same reason 

 specified above, 3 n 3 =: the sum of the numeral co-efficients of each 

 combination. 



Next, let there be four sets, taking two at a time, and one letter from 

 each as before. 



The number of possible arrangements of four sets taking two at a 



4. 3 



time == = 6 ; according to (2), each of these six arrangements 



I. • -*-■ 



produces n* combinations ; the total number = 6n' = also the sum 



of the ntimeral co-efficients. 



5. 4 

 Next, let there be five sets ; then— 1— =10 arrangements ; each pro- 



duces n % combinations, the total number of combinations = 10 n % = 

 also S. of N. C. 



6. 5 

 Next, let there be six sets, &c. &c. then = 15 = number of 



arrangements ; each produces w 2 combinations, total number of com- 

 binations =■ 15 n 2 =. S. of N. C. 



Next, let there be m sets of n things : then m = number of 



1. 2 



arrangements, the number of combinations = m n 2 = also S. 



1. 2 

 ofN. C. 



