Number of ways of choosing seven children from a classroom of 32 (15 boys, 17 girls) with at least 1 boy

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.

Division by 2 is uncorrect, as an alternative you could calculate it as

$$\binom {15}{1}\binom {17}{6}+\binom {15}{2}\binom {17}{5}+\binom {15}{3}\binom {17}{4}+\binom {15}{4}\binom {17}{3}+\binom {15}{5}\binom {17}{2}+\binom {15}{6}\binom {17}{1}+\binom {15}{7}\binom {17}{0}=\sum_{k=1}^7\binom{15}{k}\binom{17}{7-k}$$

showing also that by counting principle

$$\sum_{k=1}^7\binom{15}{k}\binom{17}{7-k}=\binom{32}{7} - \binom{17}{7}$$

and more in general the Vandermonde's identity

$$\binom{m+n}{r}=\sum_{k=0}^r\binom{m}{k}\binom{n}{r-k}$$

Firstly, dividing by $2!$ is not correct, since the groups are not indistinguishable (the first group has one boy, whereas the second group has $6$ children).

But the numerator is also not correct, since some of the resulting choices are counted exactly once, and some are counted more than once.

For example . . .

If after choosing one boy, the other $6$ choices are girls, the numerator counts that selection exactly once.

On the other hand, if boy #$1$ is chosen first, and boy #$2$ is one of the other $6$ choices, the numerator counts that selection more than once, since you can get the same result by choosing boy #$2$ first, and boy #$1$ as one of the other $6$ choices.