Let A be a set containing n elements A subset P of the set A is chosen at random. The set A is reconstructed by replacing the elements of P, and another subset Q of A is chosen at random. The probability that `P nn Q` contains exactly `m(m lt n)` elements is

Options

A. `(3^(n-m))/(4^(n))`
B. `(.^(n)C_(m)xx3^(m))/(4^(n))`
C. `(.^(n)C_(m)xx3^(n-m))/(4^(n))`
D. none of these

Answer

C

Explanation

We know that the number of subsets of a set containing n elements is `2^(n)`. Therefore, the number of subsets of a set containing n elements is `2^(n)`. Therefore, the number of ways of choosing P and Q is `.^(2n)C_(1)xx .^(2n)C_(1)=2^(n)xx 2^(n)=4^(n)`.
Out of n elements, m elements can be chosen in `.^(n)C_(m)` ways. If `P cap Q` contains exactly m elements, then from the remaining n-m elements either an element belongs to P or Q but not both P and Q. Suppose P contains r elements from the remaining (n-m) elements. Then, Q may contain any number of elements from the remaining (n-m)-r elements. Therefore, P and Q can be chosen in `.^(n-m)C_(r ) 2^((n-m)-r)`
But , r can vary from 0 to (n-m).
So, the number of ways in which P and Q can be chosen, is
`{underset(r=0)overset(n-m)sum .^(n-m)C_(r )2^((n-m)-r)}.^(n)C_(m)=(1+2)^(n-m) .^(n)C_(m)= .^(n)C_(m) 3^(n-m)`
Hence, required probability `=(.^(n)C_(m)xx3^(n-m))/(4^(n))`