Let Universe has size n.Then set of disjoint sets will be {) and U.pdf
- 1. Let Universe has size n. Then set of disjoint sets will be {) and U. Next would be if S consist of one element, then T can be all subsets formed from remaining n-1 elements. Hence every subset from one such T will make with S two disjoint sets. No of subsets when S contains one element = (n)2n-1 When S contains 2 elements, then T is set of all subsets formed from remaining n-2 Hence no of subsets = nC2(2n-2) and so on. This continues till S has all n elements and T is null set. Hence total no of disjoint sets = 1+n(2)n-1+nC2(2)n-2+...+nCn =(1+2)n = 3n ================================================== b) If both have n/2 mean no of sets =nC(n/2) 2n/2 ---------------------------------------------------------------------------------------------------- c) If Both S and T have n/3 then n/3 can be selected for S in nC(n/3) ways and for T n/3 elements can be selected from remaining 2n/3 in 2n/3 C (n/3) ways Hence No of disjoint subsets S and T with n/3 elements each = nC(n/3)2n/3 C (n/3) Solution Let Universe has size n. Then set of disjoint sets will be {) and U. Next would be if S consist of one element, then T can be all subsets formed from remaining n-1 elements. Hence every subset from one such T will make with S two disjoint sets. No of subsets when S contains one element = (n)2n-1 When S contains 2 elements, then T is set of all subsets formed from remaining n-2 Hence no of subsets = nC2(2n-2) and so on. This continues till S has all n elements and T is null set. Hence total no of disjoint sets = 1+n(2)n-1+nC2(2)n-2+...+nCn
- 2. =(1+2)n = 3n ================================================== b) If both have n/2 mean no of sets =nC(n/2) 2n/2 ---------------------------------------------------------------------------------------------------- c) If Both S and T have n/3 then n/3 can be selected for S in nC(n/3) ways and for T n/3 elements can be selected from remaining 2n/3 in 2n/3 C (n/3) ways Hence No of disjoint subsets S and T with n/3 elements each = nC(n/3)2n/3 C (n/3)