Hi 3Newton,

This is probably not a GRE question. Could you share the source?


Let us assume that there are 10 students and 10 places. We can break down the problem into 3 parts:

1. Place student A in any position.
2. Find number of options for student B
3. Arrange remaining 8 students

So let student A in position 1:

Number of options for student B = 9 (every other position is open)

Remaining 8 students: $8!$

So total number of combination = $9 \times 8!$.

Let student B in position 2:

Number of options for student B = 8 (every other position is open except the position 1 and position 2 is occupied by 2)

Remaining 8 students: $8!$

So total number of combination = $8 \times 8!$.


Now repeat this for all postions till student A in postion 9 and student B is 10.


Hence summing up all the options = $9 \times 8! + 8 \times 8! ....... 1 \times 8! = 45 \times 8! = 5\times 9 \times 8! = 5 \times 9! = \frac{10!}{2}$.

Hence option B is correct!