16 players $P_{1}, P_{2}, P_{3} ...... P_{16}$ take part in a tennis tournament. Lower suffix player is better than any higher suffix player. These players are divided into 4 groups each comprising of 4 players and the best from each group is selected for semifinals then

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Number of ways in which 16 players can be divided into four equal groups is $\frac{35}{27} \prod_{r = 1}^{8} (2 r - 1)$

Number of ways in which they can be divided into four equal groups if players $P_{1}, P_{2}, P_{3}$ and $P_{4}$ are in different groups is $\frac{11!}{108}$

Number of ways in which these 16 players can be divided into four equal groups such that when the best player is selected from each group, $P_{6}$ is one among them is $20. \frac{12!}{{(4!)}^{3}}$

Number of ways in which 16 players can be divided into four equal groups is $\frac{35}{54} \prod_{r = 1}^{8} (2 r - 1)$

answer is A, B, C.

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Detailed Solution

1) Number of ways = $\frac{16!}{{(4!)}^{4} 4!} = \frac{2^{8} .8! ∐_{r = 1}^{8} (2 r - 1)}{{(4!)}^{5}} = \frac{35}{27} \prod_{r = 1}^{8} (2 r - 1)$
2) Question Image
3)
To the ways= $= \frac{10!}{3! 7!} . \frac{12!}{{(4!)}^{3} .3!}$