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 Q of A is again chosen, the number of ways of choosing so that `(P cup Q)` is a proper subset of A, is

Options

A. `3^(n)`
B. `4^(n)`
C. `4^(n)-2^(n)`
D. `4^(n)-3^(n)`

Answer

D

Explanation

Let `A={a_(1),a_(2),a_(3), . .,a_(n)}`
a general element of A must satisfy one of the following possibilities.
[here, general element be `a_(i)(1leilen)]`
(i) `a_(i) in P,a_(i) in Q`
(ii) `a_(i) in P, a_(i) in Q`
(iii) `a_(i) in P, a_(i) in Q`
(iv) `a_(i) in P, a_(i) in Q`
Therefore, for one element `a_(i)` of A, we have four choices (i), (ii), (iii) and (iv).
`therefore`Total number of cases for all elements=`4^(n)`
and for one element `a_(i)` of A, such that `a_(i) in P cupQ,` we have three choices (i), (ii) and (iii).
`therefore`Number of cases for all elements belong to `P cup Q=3^(n)`
Hence, number of ways in which atleast one element of A does not belong to
`P cup Q=4^(n)-3^(n)`.