442 



then directly and by (13) 



(a) C(k,o,p) = 1 p = 0, 1, 2, 



A- = 0, 1, 2, 



C(ay«,1) =0 n = 1, 2, 3, ... . 



making use of (10) we obtain 



(b) C{k,n,p) = C(k,n,p—1) + (k+n—p—1) C(k,n—l,p—l) 



The left side vanishes when p = I; therefore 

 C(k,n,0) = —(k+n) C(k,n— -1,0) 

 By repeating this (n — 1) times and noting that C(k, 0,0) = 1. we obtain 



k = 0. 1, 2, 



(c) C(fc,n,0) = (— 1)"(*+1) (fc+2) .... (t+n) 



Setting p = 2. 3. 4 il, in (b), we find 



(d) 



n = 1,2, 3, 



C{k,n,p) = 



= A- r ' 



for p = 1, 2, 3. . 

 when p = 7! + 1 



Therefore for all values of k = 0, 1, 2, ; and ra = 1, 2, 3. 



(14) 2 (— l) 1 ' A(fr, k+n— i) B(i,k+n—p) = (— l) B (fc+lj (k+2) .... (it+n) 



when p = 



= when p = 1. 2. 3 . 

 = k" when p = n+1 



Example illustrating (14) for k = 2, n = 3. 





i = 



j=l 



7 = 2 



i = 3 







A(2,5-i) 



15 



— 7 



3 



—1 



sums of products 



p = 



B(i,5) 



1 



15 



85 



225 



(— 1) 3 34'5 



p = l 



B(i,A) 



1 



10 



35 



50 







p = 2 



B(i,S) 



1 



6 



11 



6 







p = 3 



B{i,2) 



1 



3 



2 











p = 4 



B{i,\) 



1 



1 











2 3 



In particular, when p = n. 



2. (— \) 1 A(k, k+n—i) B{i, k) = 



i=0 



