A bag contains `n` white and `n` red balls. Pairs of balls are drawn without replacement until the bag is empty. Show that the probability that each pair consists of one white and one red ball is `(2^n)/(^(2n)C_n)`

Options

A. `1//^(2n)C_(n)`
B. `2n//^(2n)C_(n)`
C. `2n//n!`
D. `2n//(2n!)`

Answer

B

Explanation

The required probability is
`(n^(2))/(""^(2n)C_(2))((n-1)^(2))/(""^(2n-2)C_(2))((n-2)^(2))/(""^(2n-4)C_(2))...(2^(2))/(""^(4)C_(2))(1^(2))/(""^(2)C_(2))`
`=((1xx2xx3xx4xx...xx(n-1)n)^(2))/(((2n)!)/2^(n))=(2^(n)(n!)^(2))/((2n)!)=(2^(n))/(""^(2n)C_(n))`