If there are three sections, X, Y, Z, there are a number of different ways to choose 5 questions
X Y Z
2 2 1
2 1 2
1 2 2
3 1 1
1 3 1
1 1 3
The number for the first three is
\[ 3{5 \choose 2}{5 \choose 2}{5 \choose 1} = 3(10)(10)(5) = 1,500 \]
and the number for the second three is
\[ 3{5 \choose 3}{5 \choose 1}{5 \choose 1} = 3(10)(5)(5) = 750 \]
giving a total of \( 2,250 \).
I think the problem with DarthSid's method is there is double (or more) counting. For example, call the questions
x1,x2,x3,x4,x5
y1,y2,...,y5
z1,...,z5.
Then one of the options under DS's algorithm is
1. Select one from each of x,y,z: let's take x1,y1,z1.
2. Now two of the remaining: let's take x2,y2
Hence the selection is
x1,x2,y1,y2,z1
But we can generate that same selection by
1. taking x2,y2,z1
2. then x1,y1
or
1. x1,y2,z1
2. then x2,y1