A is a set containing n elements. A subset `P_1` of A is chosen. The set A is reconstructed by replacing the elements P Next, a of subset `P_2` of A is chosen and again the set is reconstructed by replacing the elements of `P_2`, In this way, m subsets `P_1, P_2....,P_m` of A are chosen. The number of ways of choosing `P_1,P_2,P_3,P_4...P_m`

Explanation

Let `A={a_(1),a_(2),a_(3), . ..,a_(n)}`
For each `a_(i)(1leilen),` we have either `a in P_(j)` or `a_(i) cancel(in)P_(j)(1 le j le m)`. i.e., there are `2^(m)` choices in which `a_(i)(1leilen)` may belong to the `P_(j)`'s.
Out of these, there is only one choice, in which `a_(i) in P_(j)` for all `j=1,2, . .. ,m` which is not favourable for
`P_(1)capP_(2)capP_(3)cap . ..capP_(m)` to be `phi` thus,
`a_(i) cancel(in)P_(1)capP_(2)cap . ..capP_(m)` in `(2^(m)-1)` ways. since, there are n elements in the set A, the total number of choices is `(2^(m)-1)`.