Question 473509
Let's say there are 2 students A and B



Let's focus on how many ways to assign 3 prizes to student A.



There are 9*8*7 = 504 different ways to assign 3 prizes (out of 9) to student A where order matters. Since order does NOT matter, we divide by 3! = 6 to get 504/6 = 84. Note: this is to ignore combinations that have been counted more than once.



So there are 84 ways to assign 3 prizes to student A where order does NOT matter.



Now do the same work for student B to get 84 different ways as well. Basically repeat the above, but replace "Student A" with "Student B".




So there are 84+84=168 different ways to assign 3 prizes to one student and 6 prizes to the other.