Eight students are shortlisted to be prize recipients for a competition. Two students are awarded the first prize and another two students are awarded the second prize. The eight shortlisted students are to stand in a single row for a photo-taking session with a guest of honour.

1. Identify the number of distinct units to be arranged.

By grouping the two first prize awardees as $1$ unit and the two second prize awardees as another $1$ unit, the remaining $4$ students act as individual units.
Total units to arrange $= 1 + 1 + 4 = 6$ units.

2. Calculate the total number of arrangements by multiplying the permutations of the units by the internal permutations of the groups.

Number of ways to arrange the $6$ units $= 6!$
Number of ways to arrange the first prize awardees within their group $= 2!$
Number of ways to arrange the second prize awardees within their group $= 2!$

Total possible arrangements $= 6! \times 2! \times 2!$
$= 720 \times 2 \times 2$
$= 2880$

1. Fix the guest of honour to resolve the circular permutation and identify the available seats.

Fix the guest of honour at one seat. To account for the circular arrangement, there is $1$ way to do this since the $10$ seats are identical.
This leaves $9$ remaining seats: $4$ on the left, $4$ on the right, and $1$ directly opposite the guest of honour.

2. Determine the position of the empty seat and arrange the students.

Since exactly $4$ students must be seated on each side (left and right) of the guest of honour, they must occupy all $4$ seats on the left and all $4$ seats on the right. The seat directly opposite must therefore be the single empty seat.
The $8$ students can be arranged in these $8$ specific seats in $8!$ ways.
$8! = 40320$