There are more quadruples with product greater than 2009. If the elements of $\{a,b,c,d\}$ can be paired off into two pairs, each of which has product $48$, then the product of all $4$ elements is $48^2=2304>2009$. Otherwise, pair off $\{a,b,c,d\}$ with $\{\frac{48}{a},\frac{48}{b},\frac{48}{c},\frac{48}{d}\}$. The product of all $8$ elements in both quadruples is $48^4$, so one of these quadruples has product of elements $\ge 48^2=2304>2009$. Therefore, the number of quadruples that have product larger than $2009$ is at least the number of the first kind of quadruple plus half the number of quadruples of the second kind, which is strictly greater than half of (the number of the first kind + the number of the second kind), which is half the number of quadruples. Cheers, Rofler