TITLE: Number of ways of choosing seven children from a classroom of 32 (15 boys, 17 girls) with at least 1 boy
QUESTION [18 upvotes]: 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.

REPLY [5 votes]: Just for fun here is a way to modify your approach to get the correct answer as suggested by @ArsenBerk in a comment.
Assuming you want to choose the $1$ boy first in $\binom{15}{1}$ ways, then we must break the remaining choices from $31$ students into cases where $k=0,1,\ldots, 6$ remaining boys from $14$ are chosen and $6-k$ girls are chosen from $17$. For each $k$ this may be done in $\binom{14}{k}\binom{17}{6-k}$ ways. 
However each of these cases must be divided by the number of equivalent ways that $1$ of the $k+1$ chosen boys can be selected first, there are clearly $\binom{k+1}{1}$ of these so dividing each term by this, summing and multiplying this sum by the initial $\binom{15}{1}$ term gives:
$$\binom{15}{1}\sum_{k=0}^{6}\binom{14}{k}\binom{17}{6-k}/\binom{k+1}{1}\tag{Answer}$$
which you can confirm equals $\binom{32}{7}-\binom{17}{7}$. 
In fact by taking $\binom{15}{1}$ inside the summation a little algebraic manipulation shows this is just part of the Vandermonde identity summation presented by @gimusi.