A is a set containing n elements. A subset P of A is chosen. The set A is reconstructed by replacing the elements of P. A subset of A is again chosen. Find the number of ways of choosing P and Q,
so that
(i) `P capQ` contains exactly r elements.
(ii) `PcapQ` contains exactly 2 elements.
(iii) `P cap Q=phi`

Explanation

Let `A={a_(1),a_(2),a_(3), . . .,a_(n)}`
(i) The r elements in P and Q such that `PcapQ` can be chosen out of n is `.^(n)C_(R)` ways a general element of A must satisfy one of the following possibilities [here, general element be `a_(i)(1leilen)]`
(i) `a_(i)inP and a_(i) in Q`
(ii) `a_(i) in P and a_(i) cancel(in)Q`
(iii) `a_(i) in P and a_(i) in Q`
(iv) `a_(i) cancel(in)P and a_(i) cancel(in)Q`
Let `a_(1),a_(2), . . ,a_(r) in P capQ`
There is only one choice each of them (i.e., (i) choice) and three choices (ii), (iii) and (iv) for each of remaining (n-r) elements.
Hence, number of ways of remaining elements=`3^(n-r)`
Hence, number of ways in which `P capQ` contains
exactly r elements`=^(n)C_(r)xx3^(n-r)`
(ii) Put r=2, then `.^(n)C_(2)xx3^(n-2)`
(iii) Put r=0, then `.^(n)C_(0)xx3^(n)=3^(n)`.