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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 toA. `7/15`B. `14/15`C. `2/15`D. `4/15` |
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Answer» Correct Answer - B 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` |
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