ksharma12 wrote:
But that yields the same answer as 4C2? How do you solve use combinations or perm?
That comb above does not take into account the empty seat.
Here order is important, therefore, we'll use Perm.
All the possible ways in which 2 stud can be seated on 4 seats = 4P2.
Suppose, two students are always seated together. Then we'll treat them as 1 stud & the seats left to be filled are 3. Therefore no. of ways in which 2 stud are always seated together = 3P2.
Now, the no. of ways in which 2 students have a gap in between = total no. of ways - no. of ways in which students are seated together = 4P2-3P2 = 12-6 = 6