Given a set A of N people, it has 2N accessible subsets. This is not difficult to see, back we can anatomy anniversary accessible subset by artlessly allotment for anniversary aspect of A one of two possibilities: whether to accommodate that element, or not.
However, this includes the (one) abandoned set, and N singletons, which are not appropriately subgroups. So 2N − N − 1 subsets remain, which is exponential, like 2N.
However, this includes the (one) abandoned set, and N singletons, which are not appropriately subgroups. So 2N − N − 1 subsets remain, which is exponential, like 2N.
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