A 'Simple' Problem!

In a simpler case let's try to understand how many arrangements of 5 persons can be made where A, a particular is always ahead of B. Since the mutual arrangements of A and B can be in only two ways-A is ahead of B or B is ahead of A, and as both the incidences are equally probable, the number of required arrangements = $\frac{5!}{2}$

If another condition was allowed with this one, like another person C is to be ahead of D. Then the problem can be solved by the same method. Of the above arrangements in 60 ways which satisfy the first condition , 50% would have C ahead of D and the rest- D ahead of C. Then the answer would be 30.

In this way, in the given problem 10 persons have such conditions where $S_i>G_i$ and 5 such pairs are there, so it would be 3.125% of total arrangements, i.e of 10!. Thus answer = $\frac{10!}{32}=113400$

Hope this helps. :)

Easier way to imagine this is we put 10 chairs in a straight line and let them choose their own PAIRWISE-TURN (as opposed to their own INDIVIDUAL TURN). After everyone took their seats, the interviewer could readjust the turns by making the parent-child pairs to exchange their seats with each other if any of the guardian took an earlier turn than their student offspring.

Answer
= 10! / (2!)^5
= 10! / 2^5
= 10! / 32
= 113400