r:Let the given statement be defined asP(n): The number of subsets of a set containing n distinctelements=2n, for all nϵN.Step1: For n=1,L.H.S=As, the subsets of the set containing only 1 element are:Φ and the set itselfi.e.thenumber of subsets of a set containing only element=2R.H.S=21=2As, LHS=RHS, so, it is true for n=1.Step2: Let the given statement be true for n=k.P(k): The number of subsets of a set containing k distinctelements=2kNow, we need to show P(k+1) is true whenever P(k) is true.P(k+1):Let A={a1, a2, a3, a4,…, ak, b} so that A has (k+1) elements.So the subset t of A can be divided into two collections:first contains subsets of A which don t have b in them andthe second contains subsets of A which do have b in them.First collection: { },{a1},{a1, a2},{a1, a2, a3},…,{a1, a2, a3, a4,…, ak} andSecond collection: {b},{a1,b},{a1,a2,b },{a1,a2,a3,b},…,{a1,a2,a3,a4,…,ak, b}It can be clearly seen that:The number of subsets of A in first collection=The number of subsets of set with k elements i.e. {a1, a2, a3, a4,…, ak}=2kAlso it follows that the second collection must havethe same number of the subsets as that of the first = 2kSo the total number of subsets of A=2k+2k=2k+1Thus, by the principle of mathematical induction P(n) is true.