Question 1168539
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It is the same as the number of ways a 1st,2nd,and 3rd prize can be assigned to 6 different people (i.e. we can ignore the three fourth place prizes, as they simply fall into place after the first three prizes are given):

First choose 3 people from 6:  6C3 = 6!/((6-3)!3!) = 6*5*4/(3*2) = 20 ways to do this
Now assign {1st,2nd,3rd} prizes:  3P3 = 3! = 6  ways PER selection of 3 people.

Multiply 20 selections * 6 arrangements:  
20*6 = {{{highlight(120)}}}ways