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 lt j`. Assuming that the players are paired at random in each round, what is the probability that the player `P_(4)` reaches the final?

Answer

C, D

Explanation

The number of ways in which `P(1),P_(2)….,P_(8)` can be paired in four pairs
`=(1)/(4!)[(""^(8)C_(2))(""^(6)C_(2))(""^(4)C_(2))(""^(2)C_(2))]`
`=(1)/(4!)xx(8!)/(2!6!)xx(6!)/(2!6!)xx(4!)/(2!2!)xx1`
`=(1)/(4!)xx(8xx7)/(2!xx1)xx(6xx5)/(2!xx1)xx(4xx3)/(2!xx1)=(8xx7xx6xx5)/(2*2*2*2)=105`
Now, atleast two players certainly reach the second round between `P_(1),P_(1) and P_(3) and P_(4)` can reach in final if exactly two players play against each other between `P_(1),P_(2),P_(3)` and remaining player will play against one of the players from `P_(5),P_(6),P_(7),P_(8) and P_(4)` plays against one of the remaining three from `P_(5)...P_(8)`.
This can possible in
`""^(3)C_(2)xx""^(4)C_(1)xx""^(3)C_(1)=3*4*3=26` ways
`therefore` Probability that `P_(4)` and exactly one of `P_(5)...P_(8)` reach second round `=(36)/(105)=(12)/(35)`
IF `P_(1),P_(i),P_(4) and P_(i),` where i = 2 or 3 and j = 5 or 6 and 7 reach the second round, then they can be paired in 2 pairs in `(1)/(2!)("'^(4)C_(2))(""^(2)_(2))=3` ways. But `P_(4)` will reach the final, if `P_(1)` plays against `P_(i) and P_(4)` plays against `P_(j)`.
Hence, the probability that `P_(4)` will reach the final round from the second `=(1)/(3)`
`therefore` Probability that `P_(4)` will reach the final is `(12)/(35)xx(1)/(3)=(4)/(35)`.