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!

_________________

Sandy

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