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Eight players P 1 , P 2 , . . . . . . . . . . . . . P 8 play a knock – out tournament It is known that whenever the players P i and P j play, the player P i will win if i 4 reaches the final?

Eight players P1, P2, . . . . . . . . . . . . . P8 play a knock – out tournament It is known that whenever the players Pi and Pj play, the player Pi will win if i 4 reaches the final?

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
7 years ago
Hello Student,
Please find the answer to your question
The number of ways in which P1, P2 . . . . . . . . . . . . . . . . . . . , P8 can be paired in four pairs.
= 1/4! x 8 C2 x 6 C2 x 8 C2 x 4 C2 x 2 C2 = 105
Now, at least two players certainly reach the second round in between P1, P2, and P3 and P4 can reach in final if exactly two players play against each other in between P1, P2, P3 and remaining player will play against one of the players from P5, P6, P7, P8 and P4 plays against one of the remaining three from P5, P6, P7, P8
This can be possible in 3 C2 x 4 C1 x 3 C1 = 36 ways
∴ Prob. that P4 and exactly one of P5 . . . . . . . . . . . P8 reach second round
= 36/105 = 12/35
If P1, Pi, P4 and Pj where i = 2 or 3 and j = 5 or 6 or 7 reach the second round then they can be paired in 2 pairs in
1/2! x 4 C2 x 2 C2 = 3 ways
But P4 will reach the final if P1 plays against Pi and P4 plays against Pj
Hence the prob. that P4 reach the final round from the second = 1/3.
∴ prob. that P4 reach the final is 12/35 x 1/3 = 4/35.

Thanks
Aditi Chauhan
askIITians Faculty

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