SELECTED CHAPTERS FROM ALGEBRA I. R. Shafarevich Preface

Selected chapters from algebra 77We can apply Viete's formula(12)toitby putting a i = ; i . Since all the termsof the polynomial k are products of k variables taken from a 1 , ... , a n , thenreplacing a i by ; i gives rise to a factor (;1) k , namely: k (; 1 ... ; n ) =(;1) k k ( 1 ...a n ). Hence, from (12) we obtain(13) (x ; 1 ) (x ; n )=x n ; 1 ( 1 ... n )x n;1 + +(;1) n n ( 1 ... n ):This formula expresses the coecients of the polynomial f(x) =(x; 1 ) (x; n )in terms of its roots and it is also called Viete's formula. You know its special casefor the quadratic equation: in that case there are only two polynomials 1 and 2 , 1 = 1 + 2 , 2 = 1 2 .In conclusion, consider again formula (9) for the number of subsets (or thenumber of combinations). We deduced it from the binomial formula, which was,in turn, proved in Section 3 of Chapter II using the properties of the derivative.That is a rather involved method. It would be desirable to have another proof ofthis formula based only upon combinatorial reasoning. We shall give such a proofof an even more general formula. Notice that each subset N of the set M denesa partition M = N + N where N is the complement of N. We consider a moregeneral case: an arbitrary partition M = M 1 ++M r into subsets with prescribednumber of elements: n(M 1 )=n 1 , ... , n(M r )=n r . The sequence (n 1 ...n r ) willbe called the type of the partition M = M 1 + + M r . We suppose that none ofthe sets M i is empty, i.e. that all n i > 0.Since we are dealing all the time with one and only set M where n(M) =n, itshall not always be present in our notations. Denote the number of all possible partitionsof our set M which have the prescribed type (n 1 ...n r )by C(n 1 ...n r ).Of course, we must have n 1 + + n r = n. Notice also that we are taking intoaccount the order of the sets M 1 , ... , M r . For instance, for r = 2 and given n 1and n 2 , n 1 + n 2 = n, we take the partitions M = M 1 + M 2 and M = M 2 + M 1 ,with n(M 1 ) = n 1 and n(M 2 ) = n 2 , to be dierent. Indeed, if n 1 6= n 2 thesepartitions are of dierent types. Owing to this, each partition M = M 1 + M 2 de-nes one subset M 1 (the rst one) and we have a connection with the previouslyconsidered problem: C(n 1 n 2 ) = v(Mn 1 ). In other words, for any m < n, wehave v(Mm) = C(m n ; m). We shall now derive an explicit formula for thenumber C(n 1 ...n r ). Consider an arbitrary partition M = M 1 + + M r oftype (n 1 ...n r ). Suppose that at least one of the numbers n 1 , ... , n r is dierentfrom 1. For instance, suppose that n 1 > 1 and choose an arbitrary elementa 2 M 1 . Denote by M1 0 the set of all elements of M 1 dierent from a (this isthe complement oftheset fag taken as a subset of M 1 ). Then we have the partitionM 1 = M1 0 + fag and to our partition M = M 1 + + M r correspondsa new partition M = M1 0 + fag + M 2 + + M r of type (n 1 ; 1 1n 2 ...n r ).In this way fromall partitions of type (n 1 n 2 ...n r ) we obtain all partitions oftype (n 1 ; 1 1n 2 ...n r ): the partition M = M1 0 + fag + M 2 + + M r is obtainedfrom the partition M = M 1 + + M r , where M 1 = M1 0 + fag. Moreover,one partition of type (n 1 n 2 ...n r ) gives rise to n 1 dierent partitions of type(n 1 ; 1 1n 2 ...n r ), depending on the choice of a 2 M 1 . Hence, we have(14) n 1 C(n 1 n 2 ...n r )=C(n 1 ; 1 1n 2 ...n r ):