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.

Let's enumerate boys as $B_1,B_2,...,B_{15}$ and girls as $G_1,G_2,...,G_{17}$. Then in your method, we choose one of them first, say $B_1$. Then for the remaining $6$ children, say $B_2$ is among them and the rest is $G_1,G_2,...,G_5$. But this is as same as choosing $B_2$ first and then choosing other $6$ children as $B_1,G_1,G_2,...,G_5$. Although this is only one example, you can easily see that in your method we are overcounting.
