Result 



a' a + 



a'b 



+ 



a'c -f ' 



the nth term) -f 



c'a 



+ 



c'b... (1 



3 d set, 



a" + 



b" 



+ 



c" -f &c 



1832.] Summation of Polynomial Co-efficients. 189 



Obs. The co-efficients of » a form the following progression : 



Series 1,3,6,10,15, m (m— 1 ) 



Difference . 2, 3,4,5, I. 2. 



(3) Next, let three sets combine at a time, by taking one letter from 



each at a time. This is equivalent to combining two sets together, as 



was done in the preceding section, and then combining the result with 



the third set, thus, 



1st set a + b + c &c. 



2d set a + b ' + c &c. 



. (to the nth term) -|» b'a -f b'b . . . (to 



(to the wM term), &c. = w 8 , combinations; 



= n % combinations ; 



a'af'a -f a" a'b + &c. f 



a" forms with the result rc* combinations ; so does each of the n letters 



of the third set : therefore the total number of combinations formed 



= n X n* = n 9 =? also sum of numeral co-efficients. 



Next, let there be four sets ; combine three at a time, by taking one 



letter from each of the three sets at a time. 



a -f- b &c, a' ^- b' &c, af' -\- b" &c. the combinations to be 



formed thus — a of a", a a" a'": since three sets out of four, are to be 



arranged together at a time, all the possible ways of effecting this, are 



4.3.2., 

 . — == four arrangements ; now each arrangement ac- 



cording to (3) produces n s combinations : the total number of C. in 

 this case = 4 n 3 = S. of N. C. 



If m sets be taken, m. (m — 1) (m — 2) 



7 a — Z = number of arrangements ; 



and number of combinations = m (m — 1) (m — 2) 



— w s = S. of J\. C. 



1. 2. 3. 



The co-efficients of n R form the following series : 



Series 1, 4, 10, 20, 35, 56, m (m — 1) (m— 2) 



1. 2. 3. 



21 m (m— 1 ) 



Difference <£ 1. 2. 



6 . . . . w. 



C 3 6 10... 15 ... 



( 3 4 5... i 



(5) Next, let four sets combine, taking one letter from each at a 

 time, and forming combinations consisting each of four letters. This 

 is equivalent to combining three sets, taking one letter from each at a 

 time; forming combinations of three letters each, (as in the last section,) 

 and then combining the result with the fourth set. 



