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Sixteen players S(1), S(2), S(3),…,S(16)...

Sixteen players `S_(1)`, `S_(2)`, `S_(3)`,…,`S_(16)` play in a tournament. Number of ways in which they can be grouped into eight pairs so that `S_(1)` and `S_(2)` are in different groups, is equal to

A

`((14)!)/(2^(6)*6!)`

B

`((15)!)/(2^(7)*7!)`

C

`((14)!)/(2^(7)*6!)`

D

`((14)!)/(2^(6)*7!)`

Text Solution

Verified by Experts

The correct Answer is:
A
`(a)` Required number of ways
`=("Number of ways in which" 16 "players can be divided in" `8` "couples")-("Number of ways when" S_(1) "and" S_(2) "are in the same group")`
`=(16!)/(2^(8)*8!)-((14)!)/(2^(7)*7!)=(16*15*14!)/(2*2^(7)*8*7!)-((14)!)/(2^(7)*7!)`
`=((14!))/(2^(7)*7!)[(16*15)/(16)-1]=((14)*(14)!)/(14*(6!)*2^(6))=((14!))/(2^(6)*6!)`
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