I think I'm interpreting your problem correctly. I'm interpreting it
as the simple way that 3 of the boys will get 1 prize and one of the
boys will get 2 prizes. If that's the correct interpretation, then,
Choose the lucky boy to get 2 prizes 4 ways.
Choose the 2 prizes the lucky boy is to get 5C2 = 10 ways.
Distribute the remaining 3 prizes among the other 3 boys in 3! or 6 ways.
Answer 4*10*6 = 240 ways.
[If that's not the correct interpretation and you are studying
partitions, which are a little more advanced, and allow that
the first boy can take as many prizes as he likes, and that it
may even be that he takes them all and the others get none, then
it's a more complicated problem. Let me know in the thank-you
note form below if you meant this more complicated case.
Also if this is the case, tell me whether the prizes are
distinguishable or all the prizes are exactly the same.]
Edwin
