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Question

# Eight players P1,P2,P3,..........P8 play a knockout tournament. It is known that whenever the players Pi and Pj play , the player Pi will win if i<j. Assuming that the players are paired at random in each round, what is the probability that the players P4 reaches the final?

A
4/35
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B
2/35
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C
2/21
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D
2/7
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Solution

## The correct option is C 4/35The number of ways in which P1,P2,...,P8 can be paired in four pairs=14![(8C2)(6C2)(4C2)(2C2)] ways.=14!×8!2!6!×6!2!4!×4!2!2!×1=14!×8×72!×1×6×52!×1×4×32!×1=8×7×6×52.2.2.2=105Now, atleast two pairs 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 and remaining players will play against one of the players from P5,P6,P7,P8 and P4 plays against one of the remaining three from P5...P8This can be possible in 3C2×4C1×3C1=3.4.3=36 ways.∴ probability that P4 and exactly one of P5...P8 reach second round.=36105=1235If P1,Pi,P4andPj where i=2or3 and j=5or6or7 reach the second round, then they can be paired in 2 pairs in 12!(4C2)(2C2)=3 ways.But P4 will reach the final, if P1 plays against Pi and P4 plays against Pj.Hence, the probability that P4 reach the final round from the second =13∴ the probability that P4 reach the final is 1235×13=435

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