across
1. The event in which players play against each other.
Answers
Answer:
hìiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
Explanation:
The number of ways in which P
1
,P
2
,...,P
8
can be paired in four pairs
=
4!
1
[(
8
C
2
)(
6
C
2
)(
4
C
2
)(
2
C
2
)] ways.
=
4!
1
×
2!6!
8!
×
2!4!
6!
×
2!2!
4!
×1
=
4!
1
×
2!×1
8×7
×
2!×1
6×5
×
2!×1
4×3
=
2.2.2.2
8×7×6×5
=105
Now, atleast two pairs certainly reach the second round in between P
1
,P
2
and P
3
. And P
4
can reach in final if exactly two players play against each other in between P
1
,P
2
and remaining players 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 be possible in
3
C
2
×
4
C
1
×
3
C
1
=3.4.3=36 ways.
∴ probability that P
4
and exactly one of P
5
...P
8
reach second round.
=
105
36
=
35
12
If P
1
,P
i
,P
4
andP
j
where i=2or3 and j=5or6or7 reach the second round, then they can be paired in 2 pairs in
2!
1
(
4
C
2
)(
2
C
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
reach the final round from the second
=
3
1
∴ the probability that P
4
reach the final is
35
12
×
3
1
=
35
4