In illustrations 33, the number of ways of
selecting P and Q such that P `cup` Q = A , is

Options

A. `3^(n-1)`
B. `n 3^(n)`
C. `n3 ^(n-1)`
D. `2 n ^(n-1)`

Answer

c

Explanation

If P `cup` Q conains exaclty one element , both P and Q
must be non-enpty. Thus, if P has r elements, Q must have
exactly one of these r elements and any number of elements
from among the reamaining (n-r ) elements in A, so that the
number of ways of choosing Q is `""^(r)C_(1) xx2^(n-r)` .
But, P can be chosen in `""^(n)C_(r)` ways. Therefore, P and Q can be
chosen in `""^(n)C_(1) xx""^(r)C_(1)xx2^(n-r)` , when P contains r elements. As r
can very from 1 to n . Therefore, P and Q in general can be
chosen in
`sum_(r=1)^(n) ""^(n)C_(r) xx""^(n)C_(r)xx2^(n-r)`
`=sum_(r=1)^(n) ""^(n)C_(r)r2^(n-r)`
`=sum_(r=1)^(n) (n)/(r) ""^(n-r)C_(r-1).r.2^(n-r)` "" `[because ""^(n)C_(r)= (n)/(r) ""^(n-1)C_(r - 1)]`
`nsum_(r=0)^(n) ""^(n-1)C_(r-1)2^(n-r)`
`nsum_(r=0)^(n) ""^(n-1)C_(r-1)2^((n-r)-(r-0)) = n(1 + 2)^(n+1) = n3^(n-1)`