AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

The total number of students are 50.
Number of ways if both the students are in first group is
48C18×50C30=48C18
Number of ways if both are in second group is,
48C27×20C20=48C28
The total number of ways is,
50C20×30C30=50C20
So, the required probability is,

48C18+48C28

50C20

=

48! 18!30! + 48! 28!20!

50! 20!30!

=

(20)(19)+(30)(29)

(50)(49)

=

25

49