A trust has 15 identical scholarships to hand out among 4 colleges, and every college must receive at least 2 of them. In how many ways can the scholarships be handed out?

Pay each college its compulsory 2 first. That uses 8 scholarships and leaves 7 to be split among the 4 colleges with no restriction left at all. Splitting 7 identical items among 4 recipients, zero allowed, means placing 3 dividers among 7 + 3 = 10 positions: C(10,3) = 120. Trap 816 = C(18,3) is the count when a college may be given nothing. Trap 364 = C(14,3) enforces only the weaker rule 'at least one each'. Trap 455 = C(15,3) is an off-by-one slip in the divider count.