I think I figured this out. The selection of 5 out of 8 participants is given by 8C5=8C3=8*7*6/(1*2*3)=56. This is a combination.
The arrangement of prizes is a permutation and there are 5!=120 ways of arranging 5 distinct prizes (1*2*3*4*5).
The allocation of prizes applies the permutation to each of the combinations of participants, so this gives 120 ways of allocating prizes to each of the chosen participants making 56*120 in all=6,720.
I have assumed that each of the 5 chosen participants can have one and only one prize (otherwise we'd run out of prizes).