The correct option is B 435
The number of ways in which P1, P2,⋯, P8 can be paired in four pairs
=14![(8C2)(6C2)(4C2)(2C2)]=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=105
Now, atleast two players certainly reach the second round between P1, P2, P3 and P4. P4 can reach in final if exactly two players play against each other 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⋯P8.
This 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=1235
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 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 will reach the final round from the second =13
∴ Probability that P4 will reach the final is 1235×13=435