Number of ways of choosing seven children from a classroom of 32 (15 boys, 17 girls) with at least 1 boy I know that the correct solution can be calculated as:
$$ \binom {32} {7} - \binom {17}{7}$$
But why is the following solution incorrect? (I am interested in why the following reasoning is incorrect, I realize that the two numbers are not equal):
$$ \frac{15 \binom {31} {6}}{2!} $$
The reasoning is that we first pick a boy ($15$ options) and then pick $6$ children out of $31$ remaining in an arbitrary manner. Finally divide by $2!$ since the order of the two groups does not matter.

I think the best explanation is to take a group with lets say 4 boys and 3 girls. 
With your method, any of those boys could have been the first one to be chosen, and the rest was picked as the group of six kids from the remaining 31 children. Meaning, in your expression $15 {31 \choose 6}$ every group with four boys is counted four times.
The same goes for groups with 5 boys, where every such group is counted five times. Yet you divide that number by two, as if they were counted only twice.
