Sixteen players `P_(1),P_(2),P_(3)….., P_(16)` play in tournament. If they grouped into eight pair then the probability that `P_(4)` and `P_(9)` are in different groups, is equal to

Options

A. `7/15`
B. `14/15`
C. `2/15`
D. `4/15`

Answer

B

Explanation

Total ways `=(16!)/((2!)^(8)8!)`
Number of ways in which `P_(4)` and `P_(9)` are in same groups `=(14!)/((2!)^(7)7!)`
Number of ways in which they are in different groups
`=(16!)/((2!)^(8)8!)-(14!)/((2!)^(7)7!)`
`=(14!)/((2)^(7)7!)((15.16)/(2.(8))-1)=(14.14!)/((2)^(7)7!)`
Probability `=((14.14!)/((2)^(7).7!))/((16!)/((2)^(8)8!))=(14.8.2)/(15.16)=14/15`